#include <bits/stdc++.h>
using namespace std;
int binarySearch(vector<int> arr, int l, int r, int x) {
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
    return -1;
  }
}
long long gcd(long long a, long long b) {
  if (a == 0)
    return (b);
  else
    return (gcd(b % a, a));
}
long long fastpow(long long a, long long n, long long temp) {
  if (n == 0) return (1);
  if (n == 1) return ((a * temp) % 1000000007);
  if (n & 1) temp = (temp * a) % 1000000007;
  return (fastpow((a * a) % 1000000007, n / 2, temp));
}
int main() {
  int n;
  string a;
  string ans = "";
  scanf("%d", &n);
  cin >> a;
  for (int i = 0; i < n; i++) {
    if (a[i] == '0' or a[i] == '1')
      continue;
    else if (a[i] == '4') {
      ans = ans + '2';
      ans = ans + '2';
      ans = ans + '3';
    } else if (a[i] == '6') {
      ans = ans + '5';
      ans = ans + '3';
    } else if (a[i] == '8') {
      ans = ans + '2';
      ans = ans + '2';
      ans = ans + '2';
      ans = ans + '7';
    } else if (a[i] == '9') {
      ans = ans + '2';
      ans = ans + '3';
      ans = ans + '3';
      ans = ans + '7';
    } else
      ans = ans + a[i];
  }
  sort(ans.rbegin(), ans.rend());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i];
  }
  cout << endl;
  return 0;
}

### Prompt

Generate 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;
int binarySearch(vector<int> arr, int l, int r, int x) {
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
    return -1;
  }
}
long long gcd(long long a, long long b) {
  if (a == 0)
    return (b);
  else
    return (gcd(b % a, a));
}
long long fastpow(long long a, long long n, long long temp) {
  if (n == 0) return (1);
  if (n == 1) return ((a * temp) % 1000000007);
  if (n & 1) temp = (temp * a) % 1000000007;
  return (fastpow((a * a) % 1000000007, n / 2, temp));
}
int main() {
  int n;
  string a;
  string ans = "";
  scanf("%d", &n);
  cin >> a;
  for (int i = 0; i < n; i++) {
    if (a[i] == '0' or a[i] == '1')
      continue;
    else if (a[i] == '4') {
      ans = ans + '2';
      ans = ans + '2';
      ans = ans + '3';
    } else if (a[i] == '6') {
      ans = ans + '5';
      ans = ans + '3';
    } else if (a[i] == '8') {
      ans = ans + '2';
      ans = ans + '2';
      ans = ans + '2';
      ans = ans + '7';
    } else if (a[i] == '9') {
      ans = ans + '2';
      ans = ans + '3';
      ans = ans + '3';
      ans = ans + '7';
    } else
      ans = ans + a[i];
  }
  sort(ans.rbegin(), ans.rend());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i];
  }
  cout << endl;
  return 0;
}

```