#include <bits/stdc++.h>
using namespace std;
int fact[10], n, m, x[1000], cevap[1000], at;
char ar[1000];
int main() {
  fact[0] = 1;
  scanf("%d", &m);
  for (int i = 1; i <= 10; i++) fact[i] = fact[i - 1] * i;
  n = m;
  for (int i = 1; i <= n; i++) {
    char a;
    scanf(" %c", &a);
    x[i] = a - 48;
  }
  for (int i = 1; i <= n; i++) {
    if (x[i] == 2) cevap[++at] = 2;
    if (x[i] == 3) cevap[++at] = 3;
    if (x[i] == 4) {
      cevap[++at] = 2;
      cevap[++at] = 2;
      cevap[++at] = 3;
    }
    if (x[i] == 5) cevap[++at] = 5;
    if (x[i] == 6) {
      cevap[++at] = 3;
      cevap[++at] = 5;
    }
    if (x[i] == 7) cevap[++at] = 7;
    if (x[i] == 8) {
      cevap[++at] = 2;
      cevap[++at] = 2;
      cevap[++at] = 2;
      cevap[++at] = 7;
    }
    if (x[i] == 9) {
      cevap[++at] = 2;
      cevap[++at] = 3;
      cevap[++at] = 3;
      cevap[++at] = 7;
    }
  }
  sort(cevap + 1, cevap + 1 + at);
  reverse(cevap + 1, cevap + 1 + at);
  for (int i = 1; i <= at; i++) printf("%d", cevap[i]);
  printf("\n");
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 fact[10], n, m, x[1000], cevap[1000], at;
char ar[1000];
int main() {
  fact[0] = 1;
  scanf("%d", &m);
  for (int i = 1; i <= 10; i++) fact[i] = fact[i - 1] * i;
  n = m;
  for (int i = 1; i <= n; i++) {
    char a;
    scanf(" %c", &a);
    x[i] = a - 48;
  }
  for (int i = 1; i <= n; i++) {
    if (x[i] == 2) cevap[++at] = 2;
    if (x[i] == 3) cevap[++at] = 3;
    if (x[i] == 4) {
      cevap[++at] = 2;
      cevap[++at] = 2;
      cevap[++at] = 3;
    }
    if (x[i] == 5) cevap[++at] = 5;
    if (x[i] == 6) {
      cevap[++at] = 3;
      cevap[++at] = 5;
    }
    if (x[i] == 7) cevap[++at] = 7;
    if (x[i] == 8) {
      cevap[++at] = 2;
      cevap[++at] = 2;
      cevap[++at] = 2;
      cevap[++at] = 7;
    }
    if (x[i] == 9) {
      cevap[++at] = 2;
      cevap[++at] = 3;
      cevap[++at] = 3;
      cevap[++at] = 7;
    }
  }
  sort(cevap + 1, cevap + 1 + at);
  reverse(cevap + 1, cevap + 1 + at);
  for (int i = 1; i <= at; i++) printf("%d", cevap[i]);
  printf("\n");
  return 0;
}

```