#include <bits/stdc++.h>
const int MV = 9000000;
int vis[MV];
char str[MV];
int f(int x) {
  if (x == 0)
    return 1;
  else if (x == 1)
    return 0;
  else if (x == 2 || x == 3 || x == 5 || x == 7) {
    vis[x]++;
    return 0;
  } else if (x == 4) {
    vis[3]++;
    vis[2] += 2;
    return 0;
  } else if (x == 6) {
    vis[5]++;
    vis[3]++;
    return 0;
  } else if (x == 8) {
    vis[7]++;
    vis[2] += 3;
    return 0;
  } else if (x == 9) {
    vis[7]++;
    vis[2] += 1;
    vis[3] += 2;
    return 0;
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    scanf("%s", str);
    memset(vis, 0, sizeof(vis));
    int flag = 0;
    int can = 0;
    for (int i = 0; i < strlen(str); i++) {
      if (flag == 0 && str[i] == '0') {
        continue;
      } else {
        flag = 1;
        can += f(str[i] - '0');
      }
    }
    for (int i = 9; i >= 0; i--) {
      for (int j = 0; j < vis[i]; j++) {
        printf("%d", i);
      }
    }
    puts("");
  }
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

Drazil is playing a math game with Varda.

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

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. <image> = <image>.

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, <image>

### Response

```cpp
#include <bits/stdc++.h>
const int MV = 9000000;
int vis[MV];
char str[MV];
int f(int x) {
  if (x == 0)
    return 1;
  else if (x == 1)
    return 0;
  else if (x == 2 || x == 3 || x == 5 || x == 7) {
    vis[x]++;
    return 0;
  } else if (x == 4) {
    vis[3]++;
    vis[2] += 2;
    return 0;
  } else if (x == 6) {
    vis[5]++;
    vis[3]++;
    return 0;
  } else if (x == 8) {
    vis[7]++;
    vis[2] += 3;
    return 0;
  } else if (x == 9) {
    vis[7]++;
    vis[2] += 1;
    vis[3] += 2;
    return 0;
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    scanf("%s", str);
    memset(vis, 0, sizeof(vis));
    int flag = 0;
    int can = 0;
    for (int i = 0; i < strlen(str); i++) {
      if (flag == 0 && str[i] == '0') {
        continue;
      } else {
        flag = 1;
        can += f(str[i] - '0');
      }
    }
    for (int i = 9; i >= 0; i--) {
      for (int j = 0; j < vis[i]; j++) {
        printf("%d", i);
      }
    }
    puts("");
  }
  return 0;
}

```