#include <bits/stdc++.h>
using namespace std;
int ans[65] = {0}, m = 0;
void add(int temp) {
  if (temp == 0 || temp == 1)
    return;
  else if (temp == 4) {
    ans[m++] = 2;
    ans[m++] = 2;
    ans[m++] = 3;
  } else if (temp == 6) {
    ans[m++] = 3;
    ans[m++] = 5;
  } else if (temp == 8) {
    ans[m++] = 2;
    ans[m++] = 2;
    ans[m++] = 2;
    ans[m++] = 7;
  } else if (temp == 9) {
    ans[m++] = 2;
    ans[m++] = 3;
    ans[m++] = 3;
    ans[m++] = 7;
  } else {
    ans[m++] = temp;
  }
}
bool compare(int a, int b) { return a > b; }
int main() {
  int n;
  while (cin >> n) {
    char temp;
    memset(ans, 0, sizeof(ans));
    m = 0;
    for (int i = 0; i < n; ++i) {
      cin >> temp;
      add(temp - '0');
    }
    sort(ans, ans + m, compare);
    for (int i = 0; i < m; ++i) cout << ans[i];
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 ans[65] = {0}, m = 0;
void add(int temp) {
  if (temp == 0 || temp == 1)
    return;
  else if (temp == 4) {
    ans[m++] = 2;
    ans[m++] = 2;
    ans[m++] = 3;
  } else if (temp == 6) {
    ans[m++] = 3;
    ans[m++] = 5;
  } else if (temp == 8) {
    ans[m++] = 2;
    ans[m++] = 2;
    ans[m++] = 2;
    ans[m++] = 7;
  } else if (temp == 9) {
    ans[m++] = 2;
    ans[m++] = 3;
    ans[m++] = 3;
    ans[m++] = 7;
  } else {
    ans[m++] = temp;
  }
}
bool compare(int a, int b) { return a > b; }
int main() {
  int n;
  while (cin >> n) {
    char temp;
    memset(ans, 0, sizeof(ans));
    m = 0;
    for (int i = 0; i < n; ++i) {
      cin >> temp;
      add(temp - '0');
    }
    sort(ans, ans + m, compare);
    for (int i = 0; i < m; ++i) cout << ans[i];
  }
  return 0;
}

```