#include <bits/stdc++.h>
using namespace std;
map<long long int, bool> used;
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> c;
  string s;
  cin >> s;
  for (long long int i = 0; i < n; i++) {
    c[(long long int)(s[i] - '0')]++;
  }
  long long int z = c[4];
  c[2] = c[2] + 2 * z;
  c[3] = c[3] + z;
  c[4] = 0;
  z = c[6];
  c[3] += z;
  c[5] += z;
  c[6] = 0;
  z = c[8];
  c[2] = c[2] + 3 * z;
  c[7] += z;
  c[8] = 0;
  z = c[9];
  c[3] = c[3] + 2 * z;
  c[2] += z;
  c[7] += z;
  c[9] = 0;
  while (c[7]--) {
    cout << 7;
  }
  while (c[5]--) cout << 5;
  while (c[3]--) cout << 3;
  while (c[2]--) cout << 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
map<long long int, bool> used;
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> c;
  string s;
  cin >> s;
  for (long long int i = 0; i < n; i++) {
    c[(long long int)(s[i] - '0')]++;
  }
  long long int z = c[4];
  c[2] = c[2] + 2 * z;
  c[3] = c[3] + z;
  c[4] = 0;
  z = c[6];
  c[3] += z;
  c[5] += z;
  c[6] = 0;
  z = c[8];
  c[2] = c[2] + 3 * z;
  c[7] += z;
  c[8] = 0;
  z = c[9];
  c[3] = c[3] + 2 * z;
  c[2] += z;
  c[7] += z;
  c[9] = 0;
  while (c[7]--) {
    cout << 7;
  }
  while (c[5]--) cout << 5;
  while (c[3]--) cout << 3;
  while (c[2]--) cout << 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```