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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
int a[10];
char s[105];
int main() {
  scanf("%d%s", &n, s);
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    k = s[i] - '0';
    if (k == 2 || k == 3 || k == 5 || k == 7)
      a[k]++;
    else if (k == 4) {
      a[3]++;
      a[2] += 2;
    } else if (k == 6) {
      a[5]++;
      a[3]++;
    } else if (k == 8) {
      a[7]++;
      a[2] += 3;
    } else if (k == 9) {
      a[7]++;
      a[3] += 2;
      a[2]++;
    }
  }
  for (int(i) = 0; (i) < (int)(a[7]); (i)++) printf("7");
  for (int(i) = 0; (i) < (int)(a[5]); (i)++) printf("5");
  for (int(i) = 0; (i) < (int)(a[3]); (i)++) printf("3");
  for (int(i) = 0; (i) < (int)(a[2]); (i)++) printf("2");
  return 0;
}
