CF292-C

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, 

自己先意淫2!~9!能被何数给表示(尽可能的使位数多)

可以发现

F(2)=F(2)

F(3)=F(3)

F(4)=F(322)

F(5)=F(5)

F(6)=F(53)

F(7)=F(7)

F(8)=F(7222)

F(9)=F(7332)

 转换后将所有数排序输出即可。。比赛时大意地认为F(9)=F(733222),结果被hack,直降1000名。。。以后要格外认真才行

#include <iostream>
#include <string.h>
using namespace std;
char res[20];

int main()
{
    int a[10];
    int b[10];
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    int n;
    cin>>n;
    cin>>res;
    for(int i=0;i<strlen(res);i++)
    {
        a[res[i]-'0']++;
    }
    for(int i=0;i<=9;i++)
    {
        if(a[i]!=0)
        {
            switch (i)
            {
            case 2:
                b[2]+=a[i];
                break;
            case 3:
                b[3]+=a[i];
                break;
            case 4:
                b[2]+=2*a[i];
                b[3]+=a[i];
                break;
            case 5:
                b[5]+=a[i];
                break;
            case 6:
                b[5]+=a[i];
                b[3]+=a[i];
                break;
            case 7:
                b[7]+=a[i];
                break;
            case 8:
                b[7]+=a[i];
                b[2]+=3*a[i];
                break;
            case 9:
                b[7]+=a[i];
                b[3]+=2*a[i];
                b[2]+=a[i];
                break;
            }
        }
    }
    for(int i=9;i>=2;i--)
    {
        while (b[i])
        {
            cout<<i;
            b[i]--;
        }
    }
    cout<<endl;
    return 0;
}