#include <bits/stdc++.h>
using namespace std;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int cnt[10];
int ans[10];
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] - '0' == 4) {
      cnt[2] += 2;
      cnt[3]++;
    } else if (s[i] - '0' == 5 || s[i] - '0' == 7)
      cnt[s[i] - '0']++;
    else if (s[i] - '0' == 6) {
      cnt[5]++;
      cnt[3]++;
    } else if (s[i] == '8') {
      cnt[7]++;
      cnt[2] += 3;
    } else if (s[i] == '9') {
      cnt[7]++;
      cnt[3] += 2;
      cnt[2]++;
    } else
      cnt[s[i] - '0']++;
  }
  for (int i = 7; i >= 2; i--)
    while (cnt[i]--) cout << i;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int cnt[10];
int ans[10];
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] - '0' == 4) {
      cnt[2] += 2;
      cnt[3]++;
    } else if (s[i] - '0' == 5 || s[i] - '0' == 7)
      cnt[s[i] - '0']++;
    else if (s[i] - '0' == 6) {
      cnt[5]++;
      cnt[3]++;
    } else if (s[i] == '8') {
      cnt[7]++;
      cnt[2] += 3;
    } else if (s[i] == '9') {
      cnt[7]++;
      cnt[3] += 2;
      cnt[2]++;
    } else
      cnt[s[i] - '0']++;
  }
  for (int i = 7; i >= 2; i--)
    while (cnt[i]--) cout << i;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```