#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const long long lINF = 1e18;
const double EPS = 1e-12;
const int N = 10;
int a[N];
int main() {
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    char c;
    scanf(" %c", &c);
    if (c != '0' && c != '1') {
      int k = c - '0';
      for (int j = 2; j <= k; ++j) {
        v.push_back(j);
      }
    }
  }
  for (auto h : v) {
    if (h == 2 || h == 3 || h == 5 || h == 7) {
      a[h]++;
    }
    if (h == 4) {
      a[2] += 2;
    }
    if (h == 6) {
      a[2]++;
      a[3]++;
    }
    if (h == 8) {
      a[2] += 3;
    }
    if (h == 9) {
      a[3] += 2;
    }
  }
  for (int i = 0; i < a[7]; ++i) {
    printf("7");
  }
  a[2] -= 4 * a[7];
  a[3] -= 2 * a[7];
  a[5] -= a[7];
  for (int i = 0; i < a[5]; ++i) {
    printf("5");
  }
  a[2] -= 3 * a[5];
  a[3] -= a[5];
  for (int i = 0; i < a[3]; ++i) {
    printf("3");
  }
  a[2] -= a[3];
  for (int i = 0; i < a[2]; ++i) {
    printf("2");
  }
  cout << endl;
  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;
const int INF = 1e9;
const long long lINF = 1e18;
const double EPS = 1e-12;
const int N = 10;
int a[N];
int main() {
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    char c;
    scanf(" %c", &c);
    if (c != '0' && c != '1') {
      int k = c - '0';
      for (int j = 2; j <= k; ++j) {
        v.push_back(j);
      }
    }
  }
  for (auto h : v) {
    if (h == 2 || h == 3 || h == 5 || h == 7) {
      a[h]++;
    }
    if (h == 4) {
      a[2] += 2;
    }
    if (h == 6) {
      a[2]++;
      a[3]++;
    }
    if (h == 8) {
      a[2] += 3;
    }
    if (h == 9) {
      a[3] += 2;
    }
  }
  for (int i = 0; i < a[7]; ++i) {
    printf("7");
  }
  a[2] -= 4 * a[7];
  a[3] -= 2 * a[7];
  a[5] -= a[7];
  for (int i = 0; i < a[5]; ++i) {
    printf("5");
  }
  a[2] -= 3 * a[5];
  a[3] -= a[5];
  for (int i = 0; i < a[3]; ++i) {
    printf("3");
  }
  a[2] -= a[3];
  for (int i = 0; i < a[2]; ++i) {
    printf("2");
  }
  cout << endl;
  return 0;
}

```