#include <bits/stdc++.h>
using namespace std;
long long fact[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
long long divis[4];
long long p[4] = {2, 3, 5, 7};
long long pd[4][4] = {{1, 0, 0, 0}, {1, 1, 0, 0}, {3, 1, 1, 0}, {4, 2, 1, 1}};
int main() {
  int n;
  cin >> n;
  int stuff[16];
  for (int i = 0; i < n; ++i) {
    char x;
    cin >> x;
    stuff[i] = x - '0';
  }
  vector<int> decomp;
  for (int i = 0; i < n; ++i) {
    if (stuff[i] == 0 || stuff[i] == 1) continue;
    for (int k = 0; k < 4; ++k) {
      int x = fact[stuff[i]];
      int ct = 0;
      while (x % p[k] == 0) {
        ++ct;
        x /= p[k];
      }
      divis[k] += ct;
    }
  }
  for (int i = 3; i >= 0; --i) {
    while (divis[i] > 0) {
      decomp.push_back(p[i]);
      for (int p = 0; p < 4; ++p) divis[p] -= pd[i][p];
    }
  }
  sort(decomp.begin(), decomp.end());
  for (int i = decomp.size() - 1; i >= 0; --i) cout << decomp[i];
  cout << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long fact[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
long long divis[4];
long long p[4] = {2, 3, 5, 7};
long long pd[4][4] = {{1, 0, 0, 0}, {1, 1, 0, 0}, {3, 1, 1, 0}, {4, 2, 1, 1}};
int main() {
  int n;
  cin >> n;
  int stuff[16];
  for (int i = 0; i < n; ++i) {
    char x;
    cin >> x;
    stuff[i] = x - '0';
  }
  vector<int> decomp;
  for (int i = 0; i < n; ++i) {
    if (stuff[i] == 0 || stuff[i] == 1) continue;
    for (int k = 0; k < 4; ++k) {
      int x = fact[stuff[i]];
      int ct = 0;
      while (x % p[k] == 0) {
        ++ct;
        x /= p[k];
      }
      divis[k] += ct;
    }
  }
  for (int i = 3; i >= 0; --i) {
    while (divis[i] > 0) {
      decomp.push_back(p[i]);
      for (int p = 0; p < 4; ++p) divis[p] -= pd[i][p];
    }
  }
  sort(decomp.begin(), decomp.end());
  for (int i = decomp.size() - 1; i >= 0; --i) cout << decomp[i];
  cout << endl;
}

```