#include <bits/stdc++.h>
using namespace std;
int n;
char s[20];
int a[20];
int f[20][10];
int req[10];
int b[10] = {0, 2, 3, 5, 7};
int main() {
  for (int i = 1; i <= 4; i++) f[0][i] = 0;
  for (int i = 1; i <= 9; i++) {
    for (int j = 1; j <= 4; j++) f[i][j] = f[i - 1][j];
    int p = i;
    while (p % 2 == 0) {
      p = p / 2;
      f[i][1]++;
    }
    while (p % 3 == 0) {
      p = p / 3;
      f[i][2]++;
    }
    while (p % 5 == 0) {
      p = p / 5;
      f[i][3]++;
    }
    while (p % 7 == 0) {
      p = p / 7;
      f[i][4]++;
    }
  }
  scanf("%d", &n);
  gets(s);
  gets(s);
  for (int i = 1; i <= n; i++) {
    a[i] = s[i - 1] - '0';
    for (int j = 1; j <= 4; j++) req[j] += f[a[i]][j];
  }
  for (int i = 4; i >= 1; i--) {
    while (req[i]) {
      putchar(b[i] + 48);
      for (int j = 1; j <= 4; j++) req[j] -= f[b[i]][j];
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 n;
char s[20];
int a[20];
int f[20][10];
int req[10];
int b[10] = {0, 2, 3, 5, 7};
int main() {
  for (int i = 1; i <= 4; i++) f[0][i] = 0;
  for (int i = 1; i <= 9; i++) {
    for (int j = 1; j <= 4; j++) f[i][j] = f[i - 1][j];
    int p = i;
    while (p % 2 == 0) {
      p = p / 2;
      f[i][1]++;
    }
    while (p % 3 == 0) {
      p = p / 3;
      f[i][2]++;
    }
    while (p % 5 == 0) {
      p = p / 5;
      f[i][3]++;
    }
    while (p % 7 == 0) {
      p = p / 7;
      f[i][4]++;
    }
  }
  scanf("%d", &n);
  gets(s);
  gets(s);
  for (int i = 1; i <= n; i++) {
    a[i] = s[i - 1] - '0';
    for (int j = 1; j <= 4; j++) req[j] += f[a[i]][j];
  }
  for (int i = 4; i >= 1; i--) {
    while (req[i]) {
      putchar(b[i] + 48);
      for (int j = 1; j <= 4; j++) req[j] -= f[b[i]][j];
    }
  }
  return 0;
}

```