#include <bits/stdc++.h>
using namespace std;
long long dp[20];
int main() {
  memset(dp, 0, sizeof(dp));
  long long n, i, dig;
  cin >> dig;
  cin >> n;
  while (n != 0) {
    long long r = n % 10;
    if (r == 0 || r == 1) {
      n = n / 10;
      continue;
    }
    if (r >= 9) dp[3] += 2;
    if (r >= 8) dp[2] += 3;
    if (r >= 7) dp[7] += 1;
    if (r >= 6) {
      dp[2] += 1;
      dp[3] += 1;
    }
    if (r >= 5) dp[5] += 1;
    if (r >= 4) dp[2] += 2;
    if (r >= 3) dp[3] += 1;
    if (r >= 2) dp[2] += 1;
    n = n / 10;
  }
  vector<long long> v;
  while (dp[7] >= 1 && dp[5] >= 1 && dp[3] >= 2 && dp[2] >= 4) {
    v.push_back(7);
    dp[7] -= 1;
    dp[5] -= 1;
    dp[3] -= 2;
    dp[2] -= 4;
  }
  while (dp[5] >= 1 && dp[3] >= 1 && dp[2] >= 3) {
    v.push_back(5);
    dp[5] -= 1;
    dp[3] -= 1;
    dp[2] -= 3;
  }
  while (dp[3] >= 1 && dp[2] >= 1) {
    v.push_back(3);
    dp[3] -= 1;
    dp[2] -= 1;
  }
  while (dp[2] >= 1) {
    v.push_back(2);
    dp[2] -= 1;
  }
  for (i = 0; i < v.size(); i++) {
    cout << v.at(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 dp[20];
int main() {
  memset(dp, 0, sizeof(dp));
  long long n, i, dig;
  cin >> dig;
  cin >> n;
  while (n != 0) {
    long long r = n % 10;
    if (r == 0 || r == 1) {
      n = n / 10;
      continue;
    }
    if (r >= 9) dp[3] += 2;
    if (r >= 8) dp[2] += 3;
    if (r >= 7) dp[7] += 1;
    if (r >= 6) {
      dp[2] += 1;
      dp[3] += 1;
    }
    if (r >= 5) dp[5] += 1;
    if (r >= 4) dp[2] += 2;
    if (r >= 3) dp[3] += 1;
    if (r >= 2) dp[2] += 1;
    n = n / 10;
  }
  vector<long long> v;
  while (dp[7] >= 1 && dp[5] >= 1 && dp[3] >= 2 && dp[2] >= 4) {
    v.push_back(7);
    dp[7] -= 1;
    dp[5] -= 1;
    dp[3] -= 2;
    dp[2] -= 4;
  }
  while (dp[5] >= 1 && dp[3] >= 1 && dp[2] >= 3) {
    v.push_back(5);
    dp[5] -= 1;
    dp[3] -= 1;
    dp[2] -= 3;
  }
  while (dp[3] >= 1 && dp[2] >= 1) {
    v.push_back(3);
    dp[3] -= 1;
    dp[2] -= 1;
  }
  while (dp[2] >= 1) {
    v.push_back(2);
    dp[2] -= 1;
  }
  for (i = 0; i < v.size(); i++) {
    cout << v.at(i);
  }
}

```