#include <bits/stdc++.h>
using namespace std;
long long pf[10];
long long n;
long long x;
long long pw(long long base, long long p) {
  if (p == 0) return 1;
  long long hf = pw(base, p / 2);
  if (p % 2)
    return hf * hf * base;
  else
    return hf * hf;
}
int main() {
  cin >> n;
  cin >> x;
  for (int i = 0; i < n; i++) {
    long long dd = (x / pw(10, i)) % 10;
    while (dd > 1) {
      long long d = dd;
      long long j = 2;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      j = 3;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      j = 5;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      j = 7;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      dd--;
    }
  }
  string ans = "";
  long long n7 = pf[7];
  long long n5 = pf[5];
  long long n3 = pf[3];
  long long n2 = pf[2];
  pf[5] = pf[5] - pf[7];
  pf[3] = pf[3] - pf[5] - 2 * pf[7];
  pf[2] = pf[2] - 4 * pf[7] - 3 * pf[5] - pf[3];
  for (int i = 0; i < pf[7]; i++) ans += '7';
  for (int i = 0; i < pf[5]; i++) ans += '5';
  for (int i = 0; i < pf[3]; i++) ans += '3';
  for (int i = 0; i < pf[2]; i++) ans += '2';
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long pf[10];
long long n;
long long x;
long long pw(long long base, long long p) {
  if (p == 0) return 1;
  long long hf = pw(base, p / 2);
  if (p % 2)
    return hf * hf * base;
  else
    return hf * hf;
}
int main() {
  cin >> n;
  cin >> x;
  for (int i = 0; i < n; i++) {
    long long dd = (x / pw(10, i)) % 10;
    while (dd > 1) {
      long long d = dd;
      long long j = 2;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      j = 3;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      j = 5;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      j = 7;
      while (d % j == 0) {
        d = d / j;
        pf[j]++;
      }
      dd--;
    }
  }
  string ans = "";
  long long n7 = pf[7];
  long long n5 = pf[5];
  long long n3 = pf[3];
  long long n2 = pf[2];
  pf[5] = pf[5] - pf[7];
  pf[3] = pf[3] - pf[5] - 2 * pf[7];
  pf[2] = pf[2] - 4 * pf[7] - 3 * pf[5] - pf[3];
  for (int i = 0; i < pf[7]; i++) ans += '7';
  for (int i = 0; i < pf[5]; i++) ans += '5';
  for (int i = 0; i < pf[3]; i++) ans += '3';
  for (int i = 0; i < pf[2]; i++) ans += '2';
  cout << ans << endl;
  return 0;
}

```