#include <bits/stdc++.h>
using namespace std;
long long m = 1e9 + 7, num;
vector<vector<long long>> adj;
vector<long long> freq;
long long mod(long long a) {
  long long ans = (a % m + m) % m;
  return ans;
}
long long mul(long long a, long long b) {
  long long ans = mod(mod(a) * mod(b));
  return ans;
}
long long add(long long a, long long b) {
  long long ans = mod(mod(a) + mod(b));
  return ans;
}
void solve() {
  long long n, i, val, dig, count;
  string ans = "";
  cin >> n;
  cin >> val;
  unordered_map<long long, long long> m;
  while (val) {
    dig = val % 10;
    if (dig == 2) {
      m[2]++;
    } else if (dig == 3) {
      m[2]++;
      m[3]++;
    } else if (dig == 4) {
      m[2] += 3;
      m[3]++;
    } else if (dig == 5) {
      m[2] += 3;
      m[3]++;
      m[5]++;
    } else if (dig == 6) {
      m[2] += 4;
      m[3] += 2;
      m[5]++;
    } else if (dig == 7) {
      m[2] += 4;
      m[3] += 2;
      m[5]++;
      m[7]++;
    } else if (dig == 8) {
      m[2] += 7;
      m[3] += 2;
      m[5]++;
      m[7]++;
    } else if (dig == 9) {
      m[2] += 7;
      m[3] += 4;
      m[5]++;
      m[7]++;
    }
    val /= 10;
  }
  count = m[7];
  for (i = 0; i < count; i++) {
    ans.push_back('7');
    m[2] -= 4;
    m[3] -= 2;
    m[5]--;
  }
  count = m[5];
  for (i = 0; i < count; i++) {
    ans.push_back('5');
    m[2] -= 3;
    m[3]--;
  }
  count = m[3];
  for (i = 0; i < count; i++) {
    ans.push_back('3');
    m[2]--;
  }
  count = m[2];
  for (i = 0; i < count; i++) {
    ans.push_back('2');
  }
  cout << ans;
}
int main() { solve(); }

### 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;
long long m = 1e9 + 7, num;
vector<vector<long long>> adj;
vector<long long> freq;
long long mod(long long a) {
  long long ans = (a % m + m) % m;
  return ans;
}
long long mul(long long a, long long b) {
  long long ans = mod(mod(a) * mod(b));
  return ans;
}
long long add(long long a, long long b) {
  long long ans = mod(mod(a) + mod(b));
  return ans;
}
void solve() {
  long long n, i, val, dig, count;
  string ans = "";
  cin >> n;
  cin >> val;
  unordered_map<long long, long long> m;
  while (val) {
    dig = val % 10;
    if (dig == 2) {
      m[2]++;
    } else if (dig == 3) {
      m[2]++;
      m[3]++;
    } else if (dig == 4) {
      m[2] += 3;
      m[3]++;
    } else if (dig == 5) {
      m[2] += 3;
      m[3]++;
      m[5]++;
    } else if (dig == 6) {
      m[2] += 4;
      m[3] += 2;
      m[5]++;
    } else if (dig == 7) {
      m[2] += 4;
      m[3] += 2;
      m[5]++;
      m[7]++;
    } else if (dig == 8) {
      m[2] += 7;
      m[3] += 2;
      m[5]++;
      m[7]++;
    } else if (dig == 9) {
      m[2] += 7;
      m[3] += 4;
      m[5]++;
      m[7]++;
    }
    val /= 10;
  }
  count = m[7];
  for (i = 0; i < count; i++) {
    ans.push_back('7');
    m[2] -= 4;
    m[3] -= 2;
    m[5]--;
  }
  count = m[5];
  for (i = 0; i < count; i++) {
    ans.push_back('5');
    m[2] -= 3;
    m[3]--;
  }
  count = m[3];
  for (i = 0; i < count; i++) {
    ans.push_back('3');
    m[2]--;
  }
  count = m[2];
  for (i = 0; i < count; i++) {
    ans.push_back('2');
  }
  cout << ans;
}
int main() { solve(); }

```