#include <bits/stdc++.h>
using namespace std;
int bin(int a[], int n, int x) {
  int l = 1, r = n;
  int m = 0;
  while (l <= r) {
    m = (l + r) / 2;
    if (a[m] > x)
      r = m - 1;
    else if (a[m] < x)
      l = m + 1;
    else
      return m;
  }
  return l;
}
int a[1005], bo[1004], gl[1005];
int main() {
  long long n, m, i = 0, j, x, b, g, k;
  cin >> n;
  cin >> m;
  while (m) {
    x = m % 10;
    m /= 10;
    if (x == 1 || x == 0) {
      continue;
    }
    if (x == 6) {
      a[++i] = 5;
      a[++i] = 3;
    } else if (x == 4) {
      a[++i] = 3;
      a[++i] = 2;
      a[++i] = 2;
    } else if (x == 8) {
      a[++i] = 7;
      a[++i] = 2;
      a[++i] = 2;
      a[++i] = 2;
    } else if (x == 9) {
      a[++i] = 7;
      a[++i] = 3;
      a[++i] = 3;
      a[++i] = 2;
    } else
      a[++i] = x;
  }
  sort(a + 1, a + i + 1);
  while (i) {
    cout << a[i--];
  }
}

### Prompt

Your challenge is to write 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;
int bin(int a[], int n, int x) {
  int l = 1, r = n;
  int m = 0;
  while (l <= r) {
    m = (l + r) / 2;
    if (a[m] > x)
      r = m - 1;
    else if (a[m] < x)
      l = m + 1;
    else
      return m;
  }
  return l;
}
int a[1005], bo[1004], gl[1005];
int main() {
  long long n, m, i = 0, j, x, b, g, k;
  cin >> n;
  cin >> m;
  while (m) {
    x = m % 10;
    m /= 10;
    if (x == 1 || x == 0) {
      continue;
    }
    if (x == 6) {
      a[++i] = 5;
      a[++i] = 3;
    } else if (x == 4) {
      a[++i] = 3;
      a[++i] = 2;
      a[++i] = 2;
    } else if (x == 8) {
      a[++i] = 7;
      a[++i] = 2;
      a[++i] = 2;
      a[++i] = 2;
    } else if (x == 9) {
      a[++i] = 7;
      a[++i] = 3;
      a[++i] = 3;
      a[++i] = 2;
    } else
      a[++i] = x;
  }
  sort(a + 1, a + i + 1);
  while (i) {
    cout << a[i--];
  }
}

```