#include <bits/stdc++.h>
using namespace std;
int n, m, t, ret;
int a[30005];
char s[20];
int cnt[20];
int foo(int bar) {
  int ret = 0;
  return ret;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    int cif = s[i] - '0';
    for (int j = 2; j <= cif; j++) {
      if (j == 1)
        continue;
      else if (j == 4)
        cnt[2] += 2;
      else if (j == 6) {
        cnt[2]++;
        cnt[3]++;
      } else if (j == 8)
        cnt[2] += 3;
      else if (j == 9)
        cnt[3] += 2;
      else
        cnt[j]++;
    }
  }
  int cnt7 = 0;
  int cnt5 = 0;
  int cnt3 = 0;
  while (cnt[7] > 0) {
    cnt[2] -= 4;
    cnt[3] -= 2;
    cnt[5]--;
    cnt[7]--;
    cnt7++;
  }
  while (cnt[5] > 0) {
    cnt[2] -= 3;
    cnt[3]--;
    cnt[5]--;
    cnt5++;
  }
  while (cnt[3] > 0) {
    cnt[2]--;
    cnt[3]--;
    cnt3++;
  }
  for (int i = 1; i <= cnt7; i++) cout << 7;
  for (int i = 1; i <= cnt5; i++) cout << 5;
  for (int i = 1; i <= cnt3; i++) cout << 3;
  for (int i = 1; i <= cnt[2]; i++) cout << 2;
  cout << endl;
  return 0;
}

### Prompt

Your task is to create 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;
int n, m, t, ret;
int a[30005];
char s[20];
int cnt[20];
int foo(int bar) {
  int ret = 0;
  return ret;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    int cif = s[i] - '0';
    for (int j = 2; j <= cif; j++) {
      if (j == 1)
        continue;
      else if (j == 4)
        cnt[2] += 2;
      else if (j == 6) {
        cnt[2]++;
        cnt[3]++;
      } else if (j == 8)
        cnt[2] += 3;
      else if (j == 9)
        cnt[3] += 2;
      else
        cnt[j]++;
    }
  }
  int cnt7 = 0;
  int cnt5 = 0;
  int cnt3 = 0;
  while (cnt[7] > 0) {
    cnt[2] -= 4;
    cnt[3] -= 2;
    cnt[5]--;
    cnt[7]--;
    cnt7++;
  }
  while (cnt[5] > 0) {
    cnt[2] -= 3;
    cnt[3]--;
    cnt[5]--;
    cnt5++;
  }
  while (cnt[3] > 0) {
    cnt[2]--;
    cnt[3]--;
    cnt3++;
  }
  for (int i = 1; i <= cnt7; i++) cout << 7;
  for (int i = 1; i <= cnt5; i++) cout << 5;
  for (int i = 1; i <= cnt3; i++) cout << 3;
  for (int i = 1; i <= cnt[2]; i++) cout << 2;
  cout << endl;
  return 0;
}

```