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>
#include <bits/stdc++.h>
using namespace std;
const int N = 100 * 1000 + 20;
unsigned long long int m, n, a[N], b[N], i, j;
int main() {
  cin >> n >> m;
  int x = n;
  while (x--) {
    a[i] = m % 10;
    m /= 10;
    ++i;
  }
  reverse(a, a + n);
  for (int i = 0; i <= n; ++i) {
    if (a[i] == 2) {
      b[j] = 2;
      ++j;
    } else if (a[i] == 3) {
      b[j] = 3;
      ++j;
    } else if (a[i] == 4) {
      b[j] = 3;
      ++j;
      b[j] = 2;
      ++j;
      b[j] = 2;
      ++j;
    } else if (a[i] == 5) {
      b[j] = 5;
      ++j;
    } else if (a[i] == 6) {
      b[j] = 5;
      ++j;
      b[j] = 3;
      ++j;
    } else if (a[i] == 7) {
      b[j] = 7;
      ++j;
    } else if (a[i] == 8) {
      b[j] = 7;
      ++j;
      b[j] = 2;
      ++j;
      b[j] = 2;
      ++j;
      b[j] = 2;
      ++j;
    } else if (a[i] == 9) {
      b[j] = 7;
      ++j;
      b[j] = 3;
      ++j;
      b[j] = 3;
      ++j;
      b[j] = 2;
      ++j;
    } else if (a[i] == 0 || a[i] == 1) {
      ++j;
    }
  }
  sort(b, b + 100);
  reverse(b, b + 100);
  for (int i = 0; i <= 100; ++i) {
    if (b[i] != 0) cout << b[i];
  }
}
