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;
string s;
int mark[100];
vector<int> v, va, vb;
void factor(int x) {
  for (int i = 0; i < v.size(); i++) {
    if (x % v[i] == 0) {
      while (x % v[i] == 0) {
        va.push_back(v[i]);
        x /= v[i];
        mark[v[i]]++;
      }
    }
  }
}
void factor2(int x) {
  for (int i = 0; i < v.size(); i++) {
    if (x % v[i] == 0) {
      while (x % v[i] == 0) {
        mark[v[i]]--;
        x /= v[i];
      }
    }
  }
}
int main() {
  int n, m, i, j, k, x, y, z;
  cin >> n >> s;
  v.push_back(2);
  v.push_back(3);
  v.push_back(5);
  v.push_back(7);
  for (i = 0; i < s.size(); i++) {
    x = s[i] - '0';
    if (x > 1) {
      for (j = 2; j <= x; j++) factor(j);
    }
  }
  sort(va.begin(), va.end());
  for (i = va.size() - 1; i >= 0; i--) {
    if (mark[va[i]] > 0) {
      vb.push_back(va[i]);
      for (j = 2; j <= va[i]; j++) {
        factor2(j);
      }
    }
  }
  for (i = 0; i < vb.size(); i++) cout << vb[i];
  cout << endl;
  return 0;
}
