A

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.

Examples

Input

4
1234

Output

33222

Input

3
555

Output

555

Note

In the first case,

题解：此题需要知道0~9的阶乘，"0","1","2","3","322","5","35","7","2227","2337"//分别为0~9对应的阶乘相乘形式，然后笨方法，代码如下

#include<cstdio>
#include<iostream>
using namespace std;
//"0","1","2","3","322","5","35","7","2227","2337"//分别为0~9对应的阶乘相乘形式
int main()
{
    int n;
    char s[20];
    int a[100]={0};
    cin>>n;
    cin>>s;
    for(int i=0;i<n;i++)
    {
        int t=s[i]-'0';
        switch(t)
        {
            case 1:
                a[1]++;break;
            case 2:
                a[2]++;break;
            case 3:
                a[3]++;break;
            case 4:
                a[2]+=2;a[3]++;break;
            case 5:
                a[5]++;break;
            case 6:
                a[3]++;a[5]++;break;
            case 7:
                a[7]++;break;
            case 8:
                a[2]+=3;a[7]++;break;
            case 9:
                a[2]++;a[3]+=2;a[7]++;break;
        }
    }
    for(int i=9;i>=2;i--)
      for(int j=1;j<=a[i];j++)
        cout<<i;
    cout<<endl;
    return 0;
}