#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int n, long long int m) {
  if (n == 0) return 1;
  if (n == 1) return x;
  x %= m;
  if (n % 2) return (x * (power((x * x) % m, n / 2, m) % m)) % m;
  return power((x * x) % m, n / 2, m) % m;
}
int main() {
  long long int t, n;
  cin >> t >> n;
  long long int freq[10] = {0};
  while (n) {
    int x = n % 10;
    if (x == 2 || x == 3 || x == 5 || x == 7) freq[x]++;
    if (x == 4) freq[2] += 2, freq[3]++;
    if (x == 6) freq[3]++, freq[5]++;
    if (x == 8) freq[2] += 3, freq[7]++;
    if (x == 9) freq[3] += 2, freq[2]++, freq[7]++;
    n /= 10;
  }
  for (int i = 9; i >= 2; i--) {
    for (long long int j = 0; j < freq[i]; j++) cout << i;
  }
}

### Prompt

In CPP, your task is to solve 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 int power(long long int x, long long int n, long long int m) {
  if (n == 0) return 1;
  if (n == 1) return x;
  x %= m;
  if (n % 2) return (x * (power((x * x) % m, n / 2, m) % m)) % m;
  return power((x * x) % m, n / 2, m) % m;
}
int main() {
  long long int t, n;
  cin >> t >> n;
  long long int freq[10] = {0};
  while (n) {
    int x = n % 10;
    if (x == 2 || x == 3 || x == 5 || x == 7) freq[x]++;
    if (x == 4) freq[2] += 2, freq[3]++;
    if (x == 6) freq[3]++, freq[5]++;
    if (x == 8) freq[2] += 3, freq[7]++;
    if (x == 9) freq[3] += 2, freq[2]++, freq[7]++;
    n /= 10;
  }
  for (int i = 9; i >= 2; i--) {
    for (long long int j = 0; j < freq[i]; j++) cout << i;
  }
}

```