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;
int a[20] = {0};
int freq[4] = {0};
void fact(int x) {
  int f = 1;
  for (int i = 1; i <= x; i++) f = f * i;
  int prime[] = {2, 3, 5, 7};
  for (int i = 0; i < 4; i++) {
    while (f % prime[i] == 0) {
      f = f / prime[i];
      freq[i]++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 1; i <= n; i++) {
    a[i] = s[i - 1] - 48;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] > 1) fact(a[i]);
  }
  freq[2] -= freq[3];
  freq[1] -= freq[3] * 2;
  freq[0] -= freq[3] * 4;
  freq[1] -= freq[2];
  freq[0] -= freq[2] * 3;
  freq[0] -= freq[1];
  for (int i = 1; i <= freq[3]; i++) cout << 7;
  for (int i = 1; i <= freq[2]; i++) cout << 5;
  for (int i = 1; i <= freq[1]; i++) cout << 3;
  for (int i = 1; i <= freq[0]; i++) cout << 2;
}
