#include<iostream>
(720)#include<cstdio>
#include<math.h>
(865)#include<vector>
using?namespace?std;
int?main(){
????int?num=0;
????vector<int>myVector;
????int?n,up;
????cin>>n;
????for(int?i=2;i<=sqrt(n);i++){
????????int?j=i;
????????int?tempN=n;
????????int?tempNum=0;
????????vector<int>tvector;
????????while(tempN%j==0){
????????????tempNum++;
????????????tempN/=j;
????????????tvector.push_back(j);
????????????j++;
????????}
????????if(tempNum>num){
????????????num=tempNum;
????????????myVector.clear();
????????????myVector.assign(tvector.begin(),tvector.end());
????????}
????}
????if(myVector.empty()){
????????myVector.push_back(n);
????????num=1;
????}
????printf("%d\n%d",num,myVector[0]);
????for(int?i=1;i<myVector.size();i++){
????????printf("*%d",myVector[i]);
????}
}

from?math?import?sqrt
import?sys
for?s?in?sys.stdin:
????s=int(s)
????num=0
????a=[]
????for?i?in?range(2,int(sqrt(s)+1)):
????????temp=s
????????n=0
????????t=[]
????????j=0
????????p=1
????????while?True:
????????????if?temp%(p*(i+j))==0:
????????????????p*=(i+j)
????????????????t.append(i+j)
????????????????n+=1
????????????????temp//(p*(i+j))
????????????????j+=1
????????????else:
????????????????break
????????if?n>num:
????????????num=n
????????????a=t
????cout=''
????if?num>0:
????????print(num)
????????for?i?in?range(num-1):
????????????cout+=str(a[i])
????????????cout+='*'
????????cout+=str(a[-1])
????????print(cout)
????else:
????????print(1)
????????print(s)

//思路：看到樣例輸出為長(zhǎng)度和序列值，很容易想到枚舉起始值，更新記錄長(zhǎng)度。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main(){
????LL n;
????scanf("%lld",&n);
????int m=sqrt(n+10),anslen=0,ansstart;
????for(int l=2;l<=m;++l){
????????LL product=l;int r=l;?
????????while(n%product==0){
????????????product*=(++r);
????????}
????????if(r-l>anslen)
????????????anslen=r-l,ansstart=l;????????
????}
????if(anslen==0) printf("%d\n%lld",1,n);
????else{
????????printf("%d\n",anslen);
????????for(int i=0;i<anslen;++i){
????????????if(i==0) printf("%d",ansstart+i);
????????????else printf("*%d",ansstart+i);
????????}
????}????
????return 0;
}?

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int limit = (int)Math.sqrt(n);
		boolean flag = true;
		int maxlen = -1;
		int s = 0;
		for(int i = 2;i<=limit;i++){
			if(n%i==0){
				flag = false;
			}
			int start = i;
			int len = 0;
			double temp = n;
			while(temp%start==0){
				len++;
				temp /= start;
				start++;
			}
			if(len>maxlen){
				maxlen = len;
				s = i;
			}
		}
		if(flag){
			System.out.println(1);
			System.out.println(n);
		}else{
			System.out.println(maxlen);
			System.out.print(s++);
			for(int i = 1;i<maxlen;i++,s++)
				System.out.print("*"+s);
			System.out.println();
		}
	}
}

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int main() {
	int N;
	while (cin >> N) {
		//初始狀態(tài)假設(shè)為素?cái)?shù)
		int min_start = N;
		int max = 1;
		int start=2;
		int sqrtn = sqrt(N);
		for (;start <= sqrtn;++start) {//只需要循環(huán)到sqrt(N),多了沒(méi)必要，否則會(huì)運(yùn)行超時(shí)
			if (N%start == 0) {
				//找到每一個(gè)可以被N除盡的start，然后計(jì)算連續(xù)多少個(gè)數(shù)的乘積可以被N整除
				int len = 1;
				int Ncopy = N;
				Ncopy /= start;
				while (Ncopy%(++start)==0) {//為了避免計(jì)算乘積越界，我才用了除的方式
					Ncopy /= start;
					++len;
				}
				if (len > max) {
					max = len;
					start = start - len;//每次都要重置start，從原起點(diǎn)開(kāi)始
					min_start = start;
				}else if (len == max&&start < min_start) {//不是素?cái)?shù)，但是只有一個(gè)數(shù)可以除盡
					start = start - len;
					min_start = start;
				}
			}
		}
		cout << max << endl;
		for (int i = 0;i < max-1;++i) {
			cout << min_start++ << "*";
		}
		cout << min_start << endl;
	}
}

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;

int main()
{ ????int n; ????cin>>n; ????int start_min = n, len_max = 1, start; ????for(start=2;start<=sqrt(n);start++) ????{ ????????if(n%start==0) ????????{ ????????????int len = 1; ????????????int x = n, t = start; ????????????x /= start; ????????????start++; ????????????while(x%start == 0) ????????????{ ????????????????x /= start; ????????????????start++; ????????????????len++; ????????????} ????????????if(len > len_max) ????????????{ ????????????????len_max = len; ????????????????start = t; ????????????????start_min = start; ????????????}else if(len==len_max && start_min>start){ ????????????????start = t; ????????????????start_min = start; ????????????} ????????} ????} ????cout<<len_max<<endl; ????cout<<start_min; ????for(int i=1;i<len_max;i++) ????????cout<<"*"<<start_min+i; ????cout<<endl; ????return 0;
}

	

#include <iostream>
#include <cmath>
using namespace std;
int main(){
????long long n;
????cin>>n;
????long long sqr=(long long)sqrt(1.0*n);
????long long first=0,len=0; ? ? ? ? ? ? ? //len為最長(zhǎng)連續(xù)整數(shù)的長(zhǎng)度，first是其中第一個(gè)
????for(long long i=2;i<=sqr;i++){ ? ? ? ? //遍歷第一個(gè)整數(shù)
????????long long temp=1,j; ? ? ? ? ? ? ? ?//temp為當(dāng)前連續(xù)整數(shù)的乘積
????????for(j=i;;j++){
????????????temp*=j;
????????????if(n%temp!=0) break; ? ? ? ? ? //j不斷加1直到temp不能整除n
????????}
????????if(j-i>len){ ? ? ? ? ? ? ? ? ? ? ? //更新最長(zhǎng)長(zhǎng)度與第一個(gè)整數(shù)
????????????len=j-i;
????????????first=i;
????????}
????}
????if(len==0) cout<<1<<endl<<n; ? ? ? ? ?//根號(hào)n范圍內(nèi)都無(wú)解，只能輸出n本身（素?cái)?shù)）
????else{
????????cout<<len<<endl;
????????for(long long i=0;i<len;i++){
????????????cout<<first+i;
????????????if(i<len-1) cout<<"*";
????????}
????}
????return 0;
}

#include <cmath>
#include <iostream>
using namespace std;

int main()
{
	int n;
	cin >> n;
	int min_start = n;
	int max_len = 1;
	int start;
	for(start=2; start <= sqrt(n); start++){
		if(n % start == 0){
			int len = 1;
			int copy_n = n, copy_start = start;
			copy_n = copy_n / start;
			start += 1;
			while(copy_n % start == 0){
				copy_n /= start;
				start ++;
				len ++;
			}
			if(len > max_len){
				max_len = len;
				start = copy_start;
				min_start = start;
			}
			else if(len == max_len && min_start > start){
				start = copy_start;
				min_start = start;
			}
		}
	}
	cout << max_len << endl;
	cout << min_start;
	for(int i=1; i<max_len; i++)
		cout << "*" << ++min_start;
	cout << endl;
	return 0;
}

import?java.io.BufferedReader;
import?java.io.IOException;
import?java.io.InputStreamReader;

public?class?Main?{
????static?BufferedReader?br?=?new?BufferedReader(new?InputStreamReader(System.in));

????public?static?void?main(String[]?args)?{

????????int?N?=?Integer.parseInt(getLine());
????????int?maxNum?=?0,?maxIndex?=?0;
????????for?(int?i?=?2;?i?<?Math.sqrt(N);?i++)?{
????????????int?len?=?getLen(N,?i);
????????????if?(maxNum?<?len)?{
????????????????maxNum?=?len;
????????????????maxIndex?=?i;
????????????}
????????}
????????if?(maxNum?==?0)?System.out.println(1?+?"\n"?+?N);
????????else?{
????????????System.out.println(maxNum);
????????????for?(int?i?=?0;?i?<?maxNum;?i++)?{
????????????????if?(i?!=?maxNum?-?1)?System.out.print((maxIndex?+?i)?+?"*");
????????????????else?System.out.println(maxIndex?+?i);
????????????}
????????}
????}

????static?int?getLen(int?n,?int?x)?{
????????int?len?=?0;
????????for?(;?n?%?(x?+?len)?==?0;?n?=?n?/?(x?+?len++))?;
????????return?len;
????}

????static?String?getLine()?{
????????String?line?=?"";
????????try?{
????????????line?=?br.readLine();
????????}?catch?(IOException?e)?{
????????????e.printStackTrace();
????????}
????????return?line;
????}
}

#include<bits/stdc++.h>
using?namespace?std;

typedef?long?long?LL;

int?main(){
????ios::sync_with_stdio(0);
	LL?n;
	cin>>n;
	LL?m=(LL)sqrt(n*1.0),anslen=0,ansI=0;
	for(LL?i=2;i<=m;i++){
		LL?temp=1,j=i;
		while(1){
			temp*=j;
			if(n%temp!=0)?break;
			if(j-i+1>anslen){
				ansI=i;
				anslen=j-i+1;
			}
			j++;
		}
	}
	if(!anslen){
		cout<<1<<endl<<n;
	}else{
		cout<<anslen<<endl;
		for(int?i=0;i<anslen;i++){
			if(i>0)?cout<<'*';
			cout<<ansI+i;
		}
	}
	return?0;
}

#include<iostream>
#include<math.h>
using?namespace?std;

//題目求由n的因子構(gòu)成的最長(zhǎng)的連續(xù)串，并且連續(xù)串是最小的；例如630=3*5*6*7?，最長(zhǎng)連續(xù)串就是5*6*7

int?main(){
????int?i,j,n,sum,loc=1;
????int?max_count=0;//記錄最大連續(xù)長(zhǎng)度
????cin>>n;
????/*
????我的思路：
????從2~pow(i,2)【pow(i,2)<=n】?中
????第一次循環(huán)確定連續(xù)串的開(kāi)始位置
????第二次循環(huán)找滿足條件的連續(xù)串
?*/
????for(i=2;i<=sqrt(n)&&pow(i,max_count+1)<=n;i++){
????	//找到開(kāi)始位置是i的最長(zhǎng)的比之前的max_count還要大的連續(xù)字串
????????sum=1;
????????//先找到比之前的max_count多一個(gè)的連續(xù)串
????????for(j=0;j<=max_count;j++){
????????????sum?=?sum*(i+j);
????????????if(n%sum!=0)?break;
????????}
????????
????????//找到后，如果后續(xù)很能滿足條件，我接著加
????????if(j==max_count+1){
????????????max_count++;?loc=i;
????????????for(;pow(i+j,2)<=n;j++){
????????????????sum?=?sum*(i+j);
????????????????if(n%sum!=0)?break;
????????????????max_count++;
????????????}
????????????//i?=?j-1;
????????}??

????}
????
????if(loc==1){
????????cout<<"1\n"<<n;
????????//printf("1\n%d",n);?
????????return?0;
????}
????
????cout<<max_count<<endl<<loc;
????for(i=1;i<max_count;i++)
????????cout<<"*"<<i+loc;
}

#include<bits/stdc++.h>
using?namespace?std;
int?main(){
????long?long?num,beg;
????cin>>num;
????int?ma=0;
????for(long?long?i=2;i<=(sqrt(num));i++){
????????int?k=i,now=0,tp=num;
????????while(tp%k==0){
????????????now++;tp=tp/k;k++;
????????}
????????if(now>ma){
????????????ma=now;beg=i;
????????}
????}
????if(ma){
????cout<<ma<<endl;
????for(int?i=beg;i<ma+beg-1;i++)
????????cout<<i<<"*";
????cout<<ma+beg-1;}
????else{
????????cout<<1<<endl;
????????cout<<num;
????}
}

//dfs絕對(duì)會(huì)超時(shí)
//求因子數(shù)
//題目要求，先個(gè)數(shù)，再大小，因此我從小到大開(kāi)始
//思路：試除法，在得到一個(gè)除數(shù)的時(shí)候，就證明這是至少有一個(gè)因子了，就可以開(kāi)始了，接下來(lái)就要有更多
//?????連續(xù)的因子才行，如果停止了，沒(méi)有連續(xù)的因子了，就更新成最長(zhǎng)的，從小到大保證了一定是最小
//?????的連續(xù)因子。
//為了保證每個(gè)連續(xù)的因子都得到，因此需要兩層循環(huán)
#include<iostream>
#include<vector>
using?namespace?std;

int?a[2];??//由于連續(xù)，只需要存放頭尾即可，這個(gè)好厲害，看別人的代碼看到的
int?reslen=-1;

int?main(){
????int?n;
????cin>>n;
????for(int?i=2;i<=n/i;i++){
????????if(n%i==0){
????????????int?len=0;
????????????int?temp=i;
????????????while(n%temp==0){
????????????????len++;
????????????????temp*=(i+len);
????????????}
????????????if(reslen<len){
????????????????reslen=len;
????????????????a[0]=i;
????????????????a[1]=i+len-1;
????????????}
????????}
????}
????if(reslen==-1){
????????cout<<1<<endl;
????????cout<<n;
????}
????else{
????????cout<<reslen<<endl;
????????for(int?i=a[0];i<a[1];i++)?cout<<i<<"*";
????????cout<<a[1];
????}
????
????return?0;
}

/*??? 題目大意
一個(gè)整數(shù)可能存在很多個(gè)因數(shù)組合，試在所有因數(shù)組合中找出一個(gè)最大的連續(xù)序列。
*/
int maxLen=0;?? ?//最大序列長(zhǎng)度
int startPoint;?? ?//答案序列起點(diǎn)
int target;?? ?

//更新最大長(zhǎng)度以及序列起點(diǎn)
void upData(int index){
?? ?int sum = 1;
?? ?int tmpLen = 0;
?? ?int sub = 0;
?? ?int tmpIndex = index;
?? ?while (true){
?? ??? ?sum *= tmpIndex;
?? ??? ?if (sum>=target || target %sum != 0) break;
?? ??? ?tmpLen++;
?? ??? ?tmpIndex++;
?? ??? ?if (tmpLen > maxLen) {
?? ??? ??? ?startPoint = index;
?? ??? ??? ?maxLen = tmpLen;
?? ??? ?}
?? ?}
?? ?return;
}

int main(){
?? ?cin >> target;
?? ?for (int i = 2; i <= sqrt(target); i++){
?? ??? ?upData(i);
?? ?}
?? ?if (maxLen == 0) {
?? ??? ?cout << 1 << "\n" << target << endl;
?? ??? ?return 0;
?? ?}
?? ?cout << maxLen << endl;
?? ?for (int i = 0; i < maxLen; i++){
?? ??? ?cout << startPoint + i;
?? ??? ?if (i != maxLen - 1) cout << "*";
?? ?}
?? ?cout << endl;
?? ?return 0;
}
/*
筆記：
第一個(gè)循環(huán)里面，遍歷的最大值到平方根即可，到target/2可能導(dǎo)致超時(shí)
失敗原因記錄：
1.出現(xiàn)長(zhǎng)度相同的狀況:沒(méi)有輸出起點(diǎn)最小的序列組合；
2.長(zhǎng)度為0時(shí):應(yīng)輸出長(zhǎng)度為1，序列為taget本身；
*/