#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  long long t, n, i, j, k;
  cin >> n;
  string s;
  cin >> s;
  map<long long, long long> m;
  for (i = 0; i < n; i++) {
    j = s[i] - '0';
    if (j == 2 || j == 3 || j == 5 || j == 7)
      m[j]++;
    else if (j == 4) {
      m[2] += 2;
      m[3]++;
    } else if (j == 6) {
      m[3]++;
      m[5]++;
    } else if (j == 8) {
      m[7]++;
      m[2] += 3;
    } else if (j == 9) {
      m[7]++;
      m[2]++;
      m[3] += 2;
    }
  }
  for (auto it = m.rbegin(); it != m.rend(); it++) {
    j = (it->second);
    k = (it->first);
    for (i = 1; i <= j; i++) cout << k;
  }
}

### Prompt

Your task is to create 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;
int32_t main() {
  long long t, n, i, j, k;
  cin >> n;
  string s;
  cin >> s;
  map<long long, long long> m;
  for (i = 0; i < n; i++) {
    j = s[i] - '0';
    if (j == 2 || j == 3 || j == 5 || j == 7)
      m[j]++;
    else if (j == 4) {
      m[2] += 2;
      m[3]++;
    } else if (j == 6) {
      m[3]++;
      m[5]++;
    } else if (j == 8) {
      m[7]++;
      m[2] += 3;
    } else if (j == 9) {
      m[7]++;
      m[2]++;
      m[3] += 2;
    }
  }
  for (auto it = m.rbegin(); it != m.rend(); it++) {
    j = (it->second);
    k = (it->first);
    for (i = 1; i <= j; i++) cout << k;
  }
}

```