#include <bits/stdc++.h>
using namespace std;
vector<int> ans;
int cnt[105];
char s[105];
int n;
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  for (int i = (1); i <= int(n); i++) {
    if (s[i] >= '2') cnt[2]++;
    if (s[i] >= '3') cnt[3]++;
    if (s[i] >= '4') cnt[2]++, cnt[2]++;
    if (s[i] >= '5') cnt[5]++;
    if (s[i] >= '6') cnt[2]++, cnt[3]++;
    if (s[i] >= '7') cnt[7]++;
    if (s[i] >= '8') cnt[2]++, cnt[2]++, cnt[2]++;
    if (s[i] >= '9') cnt[3]++, cnt[3]++;
  }
  while (cnt[7]) {
    ans.push_back(7);
    cnt[2] -= 4;
    cnt[3] -= 2;
    cnt[5] -= 1;
    cnt[7] -= 1;
  }
  while (cnt[5]) {
    ans.push_back(5);
    cnt[2] -= 3;
    cnt[3] -= 1;
    cnt[5] -= 1;
  }
  while (cnt[3]) {
    ans.push_back(3);
    cnt[2] -= 1;
    cnt[3] -= 1;
  }
  while (cnt[2]) {
    ans.push_back(2);
    cnt[2] -= 1;
  }
  for (int i = (0); i <= int(ans.size() - 1); i++) printf("%d", ans[i]);
  return 0;
}

### Prompt

Please formulate 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;
vector<int> ans;
int cnt[105];
char s[105];
int n;
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  for (int i = (1); i <= int(n); i++) {
    if (s[i] >= '2') cnt[2]++;
    if (s[i] >= '3') cnt[3]++;
    if (s[i] >= '4') cnt[2]++, cnt[2]++;
    if (s[i] >= '5') cnt[5]++;
    if (s[i] >= '6') cnt[2]++, cnt[3]++;
    if (s[i] >= '7') cnt[7]++;
    if (s[i] >= '8') cnt[2]++, cnt[2]++, cnt[2]++;
    if (s[i] >= '9') cnt[3]++, cnt[3]++;
  }
  while (cnt[7]) {
    ans.push_back(7);
    cnt[2] -= 4;
    cnt[3] -= 2;
    cnt[5] -= 1;
    cnt[7] -= 1;
  }
  while (cnt[5]) {
    ans.push_back(5);
    cnt[2] -= 3;
    cnt[3] -= 1;
    cnt[5] -= 1;
  }
  while (cnt[3]) {
    ans.push_back(3);
    cnt[2] -= 1;
    cnt[3] -= 1;
  }
  while (cnt[2]) {
    ans.push_back(2);
    cnt[2] -= 1;
  }
  for (int i = (0); i <= int(ans.size() - 1); i++) printf("%d", ans[i]);
  return 0;
}

```