#include <bits/stdc++.h>
using namespace std;
int n;
char c, w[50];
void solve() {
  int i, j, a[50], x, y,
      s[10] = {0, 0, 2, 6, 24, 120, 720, 5040, 40320, 362880}, m[50];
  vector<int> v, h;
  gets(w);
  for (i = 0; i < 50; i++) {
    m[i] = 0;
  }
  for (i = 0; i < n; i++) {
    scanf("%c", &c);
    a[i] = int(c - '0');
  }
  for (i = 0; i < n; i++) {
    if (a[i] >= 2) {
      y = s[a[i]];
      for (j = 2; j < 10; j++) {
        x = 0;
        while (y % j == 0 && y != 0) {
          y /= j;
          x++;
        }
        if (x != 0) {
          m[j] += x;
        }
      }
    }
  }
  for (i = 0; i < 50; i++) {
    while (m[49 - i] > 0) {
      y = s[49 - i];
      for (j = 2; j < 10; j++) {
        x = 0;
        while (y % j == 0 && y != 0) {
          y /= j;
          x++;
        }
        if (x != 0) {
          m[j] -= x;
        }
      }
      v.push_back(49 - i);
    }
  }
  sort(v.begin(), v.end());
  for (i = v.size() - 1; i >= 0; i--) {
    printf("%d", v[i]);
  }
  return;
}
int main() {
  scanf("%d", &n);
  solve();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 n;
char c, w[50];
void solve() {
  int i, j, a[50], x, y,
      s[10] = {0, 0, 2, 6, 24, 120, 720, 5040, 40320, 362880}, m[50];
  vector<int> v, h;
  gets(w);
  for (i = 0; i < 50; i++) {
    m[i] = 0;
  }
  for (i = 0; i < n; i++) {
    scanf("%c", &c);
    a[i] = int(c - '0');
  }
  for (i = 0; i < n; i++) {
    if (a[i] >= 2) {
      y = s[a[i]];
      for (j = 2; j < 10; j++) {
        x = 0;
        while (y % j == 0 && y != 0) {
          y /= j;
          x++;
        }
        if (x != 0) {
          m[j] += x;
        }
      }
    }
  }
  for (i = 0; i < 50; i++) {
    while (m[49 - i] > 0) {
      y = s[49 - i];
      for (j = 2; j < 10; j++) {
        x = 0;
        while (y % j == 0 && y != 0) {
          y /= j;
          x++;
        }
        if (x != 0) {
          m[j] -= x;
        }
      }
      v.push_back(49 - i);
    }
  }
  sort(v.begin(), v.end());
  for (i = v.size() - 1; i >= 0; i--) {
    printf("%d", v[i]);
  }
  return;
}
int main() {
  scanf("%d", &n);
  solve();
  return 0;
}

```