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>
const int a[8][4] = {{1, 0, 0, 0}, {1, 1, 0, 0}, {3, 1, 0, 0}, {3, 1, 1, 0},
                     {4, 2, 1, 0}, {4, 2, 1, 1}, {7, 2, 1, 1}, {7, 4, 1, 1}};
int n;
int b[20];
int c[4];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%1d", &b[i]);
    if (b[i] <= 1) continue;
    c[0] += a[b[i] - 2][0];
    c[1] += a[b[i] - 2][1];
    c[2] += a[b[i] - 2][2];
    c[3] += a[b[i] - 2][3];
  }
  for (int i = 1; i <= c[3]; ++i) {
    printf("7");
    c[0] -= 4;
    c[1] -= 2;
    c[2] -= 1;
  }
  for (int i = 1; i <= c[2]; ++i) {
    printf("5");
    c[0] -= 3;
    c[1] -= 1;
  }
  for (int i = 1; i <= c[1]; ++i) {
    printf("3");
    c[0] -= 1;
  }
  for (int i = 1; i <= c[0]; ++i) printf("2");
  return 0;
}
