#include <bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 5;
const long long mod = 10000000000000007;
const long long nary = 9;
long long a;
int ans[100], n;
int main() {
  while (scanf("%d%I64d", &n, &a) != EOF) {
    n = 0;
    while (a > 0) {
      int d = (int)(a % 10);
      a /= (long long)10;
      if (d == 9) {
        d -= 2;
        ans[++n] = 3;
        ans[++n] = 3;
        ans[++n] = 2;
      }
      if (d == 8) {
        d--;
        ans[++n] = 2;
        ans[++n] = 2;
        ans[++n] = 2;
      }
      if (d == 6) {
        d--;
        ans[++n] = 3;
      }
      if (d == 4) {
        d--;
        ans[++n] = 2;
        ans[++n] = 2;
      }
      if (d > 1) ans[++n] = d;
    }
    sort(ans + 1, ans + n + 1);
    for (int i = n; i > 0; i--) printf("%d", ans[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 maxn = 6e5 + 5;
const long long mod = 10000000000000007;
const long long nary = 9;
long long a;
int ans[100], n;
int main() {
  while (scanf("%d%I64d", &n, &a) != EOF) {
    n = 0;
    while (a > 0) {
      int d = (int)(a % 10);
      a /= (long long)10;
      if (d == 9) {
        d -= 2;
        ans[++n] = 3;
        ans[++n] = 3;
        ans[++n] = 2;
      }
      if (d == 8) {
        d--;
        ans[++n] = 2;
        ans[++n] = 2;
        ans[++n] = 2;
      }
      if (d == 6) {
        d--;
        ans[++n] = 3;
      }
      if (d == 4) {
        d--;
        ans[++n] = 2;
        ans[++n] = 2;
      }
      if (d > 1) ans[++n] = d;
    }
    sort(ans + 1, ans + n + 1);
    for (int i = n; i > 0; i--) printf("%d", ans[i]);
    printf("\n");
  }
  return 0;
}

```