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;
const int zs[4] = {2, 3, 5, 7};
int n, v[10], f[10];
string x;
bool zs1(int z) {
  for (int i = 0; i <= 3; i++) {
    if (z == zs[i]) return true;
  }
  return false;
}
void judge(int s) {
  int i = 3;
  long long x = f[s];
  while (x > 1) {
    if (x % f[zs[i]] == 0) {
      v[zs[i]]++;
      x /= f[zs[i]];
    } else
      i--;
  }
}
int main() {
  cin >> n;
  cin >> x;
  int jc = 1;
  for (int i = 2; i <= 9; i++) {
    jc *= i;
    f[i] = jc;
  }
  int t = x.size();
  for (int i = 0; i < t; i++) {
    int z = x[i] - '0';
    if (z == 0 || z == 1) continue;
    if (zs1(z))
      v[z]++;
    else {
      judge(z);
    }
  }
  for (int i = 9; i >= 2; i--) {
    for (int j = 1; j <= v[i]; j++) cout << i;
  }
  return 0;
}
