C. Drazil and Factorial

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Drazil is playing a math game with Varda.

Let's define for positive integer x as a product of factorials of its digits. For example, .

First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:

1. x doesn't contain neither digit 0 nor digit 1.

2. = .

Help friends find such number.

Input

The first line contains an integer n (1 ≤ n ≤ 15) — the number of digits in a.

The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes.

Output

Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.

Sample test(s)

Input

4
1234

Output

33222

Input

3
555

Output

555

Note

In the first case,

根据题目给出的数据可以猜到要么sqrt（）简化数据，要么找规律；

其实就是将每个数字拆分为阶乘相乘形式，并且要求最大，由于2~9每个数字的阶乘相乘的形式是固定的，凡素数无法拆，此时答案就固定了；

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
   char str[][20]={"0","1","2","3","322","5","35","7","2227","2337"};//括号里面分别为0~9对应的阶乘相乘形式 比如6！=3！*5！
   int n;
   cin>>n;
   char s[500]={"\0"},c[20];

   cin>>c;
   for(int i=0;i<n;i++)
   {
       int m=c[i]-'0';
       if(m>1) strcat(s,str[m]);
   }

   int len=strlen(s);
   sort(s,s+len);
   for(int i=len-1;i>=0;i--)
    cout<<s[i];
   cout<<endl;
}