#include <bits/stdc++.h>
using namespace std;
const int N = 100001;
const int mod = (int)(1e9 + 7);
int n;
string s;
string ans;
int p[40];
int cnt[40];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  cin >> s;
  for (int i = 0; i < n; ++i) p[i] = s[i] - '0';
  sort(p, p + n);
  for (int i = 0; i < n; ++i) {
    for (int j = 2; j <= p[i]; ++j) cnt[j]++;
  }
  int delta4 = cnt[4] - cnt[5];
  int delta6 = cnt[6] - cnt[7];
  cnt[4] = cnt[5];
  cnt[3] += 2 * cnt[9] + delta6;
  cnt[2] += 3 * cnt[8] + delta4 * 2 + delta6;
  cnt[9] = 0;
  cnt[8] = 0;
  cnt[6] = cnt[7];
  for (int j = 2; j <= 7; ++j) {
    int x = cnt[j] - cnt[j + 1];
    for (int i = 1; i <= x; ++i) ans += char(j + '0');
  }
  reverse(ans.begin(), ans.end());
  cout << ans;
}

### Prompt

Please create a solution in CPP 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;
const int N = 100001;
const int mod = (int)(1e9 + 7);
int n;
string s;
string ans;
int p[40];
int cnt[40];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  cin >> s;
  for (int i = 0; i < n; ++i) p[i] = s[i] - '0';
  sort(p, p + n);
  for (int i = 0; i < n; ++i) {
    for (int j = 2; j <= p[i]; ++j) cnt[j]++;
  }
  int delta4 = cnt[4] - cnt[5];
  int delta6 = cnt[6] - cnt[7];
  cnt[4] = cnt[5];
  cnt[3] += 2 * cnt[9] + delta6;
  cnt[2] += 3 * cnt[8] + delta4 * 2 + delta6;
  cnt[9] = 0;
  cnt[8] = 0;
  cnt[6] = cnt[7];
  for (int j = 2; j <= 7; ++j) {
    int x = cnt[j] - cnt[j + 1];
    for (int i = 1; i <= x; ++i) ans += char(j + '0');
  }
  reverse(ans.begin(), ans.end());
  cout << ans;
}

```