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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[10];
int main() {
  int n = read();
  for (int i = 1; i <= (n); i++) {
    char ch = getchar();
    while (ch < '0' || ch > '9') ch = getchar();
    a[ch - '0']++;
  }
  for (int i = (9); i >= (1); i--) {
    if (i == 9) {
      a[3] += 2 * a[9];
      a[7] += a[9];
      a[2] += a[9];
      a[9] = 0;
    } else if (i == 8) {
      a[7] += a[8];
      a[2] += 3 * a[8];
      a[8] = 0;
    } else if (i == 6) {
      a[3] += a[6];
      a[5] += a[6];
      a[6] = 0;
    } else if (i == 4) {
      a[2] += 2 * a[4];
      a[3] += a[4];
      a[4] = 0;
    }
  }
  for (int i = (9); i >= (2); i--)
    for (int j = 1; j <= (a[i]); j++) putchar('0' + i);
  printf("\n");
  return 0;
}
