#include <bits/stdc++.h>
using namespace std;
long long fast_pow(long long base, long long n, long long m) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long halfn = fast_pow(base, n / 2, m);
  if (n % 2 == 0)
    return (halfn * halfn) % m;
  else
    return (((halfn * halfn) % m) * base) % m;
}
long long findMMI_fermat(long long n, long long m) {
  return fast_pow(n, m - 2, m);
}
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  vector<long long> fact(12, 0);
  fact[0] = 1;
  for (long long i = 1; i < 10; i++) {
    fact[i] = i * fact[i - 1];
  }
  long long num;
  cin >> num;
  string n;
  cin >> n;
  string ans;
  vector<long long> cnt(11, 0);
  for (long long i = 0; i < num; i++) {
    long long k = n[i] - '0';
    while (k > 1) {
      if (k == 4) {
        cnt[2] += 2;
      } else if (k == 6) {
        cnt[2]++;
        cnt[3]++;
      } else if (k == 8) {
        cnt[2] += 3;
      } else if (k == 9) {
        cnt[3] += 2;
      } else {
        cnt[k]++;
      }
      k--;
    }
  }
  for (long long i = 10 - 1; i >= 2; i--) {
    if (cnt[i] > 0) {
      long long k = cnt[i];
      char c = (i + '0');
      for (long long j = 0; j < k; j++) ans += c;
      if (i == 7) {
        cnt[5] -= k;
        cnt[3] -= (2 * k);
        cnt[2] -= (4 * k);
      } else if (i == 5) {
        cnt[2] -= (3 * k);
        cnt[3] -= k;
      } else if (i == 3) {
        cnt[2] -= k;
      } else {
        cnt[i] -= k;
      }
    }
  }
  sort((ans).begin(), (ans).end());
  reverse((ans).begin(), (ans).end());
  cout << ans << '\n';
}

### 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;
long long fast_pow(long long base, long long n, long long m) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long halfn = fast_pow(base, n / 2, m);
  if (n % 2 == 0)
    return (halfn * halfn) % m;
  else
    return (((halfn * halfn) % m) * base) % m;
}
long long findMMI_fermat(long long n, long long m) {
  return fast_pow(n, m - 2, m);
}
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  vector<long long> fact(12, 0);
  fact[0] = 1;
  for (long long i = 1; i < 10; i++) {
    fact[i] = i * fact[i - 1];
  }
  long long num;
  cin >> num;
  string n;
  cin >> n;
  string ans;
  vector<long long> cnt(11, 0);
  for (long long i = 0; i < num; i++) {
    long long k = n[i] - '0';
    while (k > 1) {
      if (k == 4) {
        cnt[2] += 2;
      } else if (k == 6) {
        cnt[2]++;
        cnt[3]++;
      } else if (k == 8) {
        cnt[2] += 3;
      } else if (k == 9) {
        cnt[3] += 2;
      } else {
        cnt[k]++;
      }
      k--;
    }
  }
  for (long long i = 10 - 1; i >= 2; i--) {
    if (cnt[i] > 0) {
      long long k = cnt[i];
      char c = (i + '0');
      for (long long j = 0; j < k; j++) ans += c;
      if (i == 7) {
        cnt[5] -= k;
        cnt[3] -= (2 * k);
        cnt[2] -= (4 * k);
      } else if (i == 5) {
        cnt[2] -= (3 * k);
        cnt[3] -= k;
      } else if (i == 3) {
        cnt[2] -= k;
      } else {
        cnt[i] -= k;
      }
    }
  }
  sort((ans).begin(), (ans).end());
  reverse((ans).begin(), (ans).end());
  cout << ans << '\n';
}

```