#include <bits/stdc++.h>
using namespace std;
int n, i, j, num, b[1000000];
int jie(int x) {
  int res = 1;
  for (int i = 1; i <= x; i++) res *= i;
  return res;
}
bool cmp(int a, int b) { return a > b; }
int main() {
  scanf("%d\n", &n);
  for (i = 1; i <= n; i++) {
    int x = getchar() - '0';
    if (x <= 1) continue;
    if (x <= 3)
      b[++num] = x;
    else if (x == 4)
      b[++num] = 2, b[++num] = 2, b[++num] = 3;
    else if (x == 5)
      b[++num] = x;
    else if (x == 6)
      b[++num] = 5, b[++num] = 3;
    else if (x == 7)
      b[++num] = x;
    else if (x == 8) {
      b[++num] = 7;
      for (j = 1; j <= 3; j++) b[++num] = 2;
    } else {
      b[++num] = 7;
      b[++num] = 3, b[++num] = 3;
      b[++num] = 2;
    }
  }
  sort(b + 1, b + num + 1, cmp);
  for (i = 1; i <= num; i++) printf("%d", b[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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, i, j, num, b[1000000];
int jie(int x) {
  int res = 1;
  for (int i = 1; i <= x; i++) res *= i;
  return res;
}
bool cmp(int a, int b) { return a > b; }
int main() {
  scanf("%d\n", &n);
  for (i = 1; i <= n; i++) {
    int x = getchar() - '0';
    if (x <= 1) continue;
    if (x <= 3)
      b[++num] = x;
    else if (x == 4)
      b[++num] = 2, b[++num] = 2, b[++num] = 3;
    else if (x == 5)
      b[++num] = x;
    else if (x == 6)
      b[++num] = 5, b[++num] = 3;
    else if (x == 7)
      b[++num] = x;
    else if (x == 8) {
      b[++num] = 7;
      for (j = 1; j <= 3; j++) b[++num] = 2;
    } else {
      b[++num] = 7;
      b[++num] = 3, b[++num] = 3;
      b[++num] = 2;
    }
  }
  sort(b + 1, b + num + 1, cmp);
  for (i = 1; i <= num; i++) printf("%d", b[i]);
  return 0;
}

```