#include <bits/stdc++.h>
using namespace std;
int n;
long long s;
int bit[15], bit2[15], bit3[15];
int main() {
  memset(bit, 0, sizeof(bit));
  memset(bit2, 0, sizeof(bit2));
  memset(bit3, 0, sizeof(bit3));
  cin >> n >> s;
  while (s) {
    int c = s % 10;
    for (int i = 2; i <= c; ++i) bit2[i]++;
    s /= 10;
  }
  for (int i = 9; i >= 2; --i)
    if (bit2[i]) {
      if (i == 9) {
        bit2[2] += bit2[9];
        bit3[3] += bit2[9] * 2;
        bit2[8] -= bit2[9];
      } else if (i == 6)
        bit3[3] += bit2[i];
      else if (i == 8)
        bit3[2] += bit2[i] * 3;
      else if (i == 4)
        bit3[2] += bit2[i] * 2;
      else {
        bit3[i] += bit2[i];
        for (int j = 2; j <= i; ++j) bit2[j] -= bit2[i];
      }
    }
  for (int i = 9; i >= 2; --i)
    for (int j = 1; j <= bit3[i]; ++j) printf("%d", i);
  puts("");
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int n;
long long s;
int bit[15], bit2[15], bit3[15];
int main() {
  memset(bit, 0, sizeof(bit));
  memset(bit2, 0, sizeof(bit2));
  memset(bit3, 0, sizeof(bit3));
  cin >> n >> s;
  while (s) {
    int c = s % 10;
    for (int i = 2; i <= c; ++i) bit2[i]++;
    s /= 10;
  }
  for (int i = 9; i >= 2; --i)
    if (bit2[i]) {
      if (i == 9) {
        bit2[2] += bit2[9];
        bit3[3] += bit2[9] * 2;
        bit2[8] -= bit2[9];
      } else if (i == 6)
        bit3[3] += bit2[i];
      else if (i == 8)
        bit3[2] += bit2[i] * 3;
      else if (i == 4)
        bit3[2] += bit2[i] * 2;
      else {
        bit3[i] += bit2[i];
        for (int j = 2; j <= i; ++j) bit2[j] -= bit2[i];
      }
    }
  for (int i = 9; i >= 2; --i)
    for (int j = 1; j <= bit3[i]; ++j) printf("%d", i);
  puts("");
  return 0;
}

```