#include <bits/stdc++.h>
using namespace std;
const int M = 1e4 + 4;
long long t[10];
char g[100000];
int main() {
  char s[20];
  int n;
  cin >> n;
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    int p = s[i] - '0';
    if (p == 4) {
      t[2] += 2;
      t[3]++;
      continue;
    }
    if (p == 6) {
      t[5]++;
      t[3]++;
      continue;
    }
    if (p == 8) {
      t[7]++;
      t[2] += 3;
      continue;
    }
    if (p == 9) {
      t[7]++;
      t[2]++;
      t[3] += 2;
      continue;
    }
    t[p]++;
  }
  int k = 0;
  for (int i = 9; i > 1; i--) {
    while (t[i]) {
      g[k++] = i + '0';
      t[i]--;
    }
  }
  g[k] = 0;
  puts(g);
}

### 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;
const int M = 1e4 + 4;
long long t[10];
char g[100000];
int main() {
  char s[20];
  int n;
  cin >> n;
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    int p = s[i] - '0';
    if (p == 4) {
      t[2] += 2;
      t[3]++;
      continue;
    }
    if (p == 6) {
      t[5]++;
      t[3]++;
      continue;
    }
    if (p == 8) {
      t[7]++;
      t[2] += 3;
      continue;
    }
    if (p == 9) {
      t[7]++;
      t[2]++;
      t[3] += 2;
      continue;
    }
    t[p]++;
  }
  int k = 0;
  for (int i = 9; i > 1; i--) {
    while (t[i]) {
      g[k++] = i + '0';
      t[i]--;
    }
  }
  g[k] = 0;
  puts(g);
}

```