#include <bits/stdc++.h>
using namespace std;
int factor[10][10] = {{0}, {1},    {2}, {3},       {2, 2},
                      {5}, {2, 3}, {7}, {2, 2, 2}, {3, 3}};
int factor_n[10] = {0, 0, 1, 1, 2, 1, 2, 1, 3, 2};
int p_num[10] = {0};
int prime[10] = {1, 1, 1, 1, 0, 1, 0, 1, 0, 0};
int ans[10] = {0};
int main() {
  int a[20];
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%1d", &a[i]);
    if (a[i] == 2 || a[i] == 3) {
      ans[a[i]]++;
    } else if (!prime[a[i]]) {
      for (int j = a[i]; j >= 2; j--) {
        if (prime[j]) {
          ans[j]++;
          break;
        }
        for (int k = 0; k < factor_n[j]; k++) {
          p_num[factor[j][k]]++;
        }
      }
    } else
      ans[a[i]]++;
  }
  ans[3] += p_num[3];
  ans[2] += p_num[2] - p_num[3];
  for (int i = 7; i > 1; i--) {
    if (ans[i] != 0) {
      for (int j = 0; j < ans[i]; j++) {
        cout << i;
      }
    }
  }
  return 0;
}

### Prompt

Generate a Cpp solution to 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>
using namespace std;
int factor[10][10] = {{0}, {1},    {2}, {3},       {2, 2},
                      {5}, {2, 3}, {7}, {2, 2, 2}, {3, 3}};
int factor_n[10] = {0, 0, 1, 1, 2, 1, 2, 1, 3, 2};
int p_num[10] = {0};
int prime[10] = {1, 1, 1, 1, 0, 1, 0, 1, 0, 0};
int ans[10] = {0};
int main() {
  int a[20];
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%1d", &a[i]);
    if (a[i] == 2 || a[i] == 3) {
      ans[a[i]]++;
    } else if (!prime[a[i]]) {
      for (int j = a[i]; j >= 2; j--) {
        if (prime[j]) {
          ans[j]++;
          break;
        }
        for (int k = 0; k < factor_n[j]; k++) {
          p_num[factor[j][k]]++;
        }
      }
    } else
      ans[a[i]]++;
  }
  ans[3] += p_num[3];
  ans[2] += p_num[2] - p_num[3];
  for (int i = 7; i > 1; i--) {
    if (ans[i] != 0) {
      for (int j = 0; j < ans[i]; j++) {
        cout << i;
      }
    }
  }
  return 0;
}

```