#include <bits/stdc++.h>
const int N = 15 + 10;
int n, k, a[10], f[10], ok = 1;
char s[N];
void work(int s, int t) {
  if (s == 2) f[2] += t;
  if (s == 3) f[3] += t;
  if (s == 4) f[2] += 2 * t;
  if (s == 5) f[5] += t;
  if (s == 6) {
    f[2] += t;
    f[3] += t;
  }
  if (s == 7) f[7] += t;
  if (s == 8) f[2] += 3 * t;
  if (s == 9) f[3] += 2 * t;
  return;
}
int main() {
  scanf("%d%s", &n, s);
  for (int i = 0; i < n; i++)
    for (int j = 2; j <= s[i] - '0'; j++) a[j]++;
  for (int i = 2; i < 10; i++)
    if (a[i]) ok = 0;
  if (ok) {
    puts("0");
    return 0;
  }
  for (int i = 2; i < 10; i++)
    while (a[i]--) work(i, 1);
  for (int i = 9; i > 1; i--)
    while (f[i]) {
      putchar(i + '0');
      for (int j = i; j > 1; j--) work(j, -1);
    }
  return 0;
}

### Prompt

Create a solution in CPP for 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>
const int N = 15 + 10;
int n, k, a[10], f[10], ok = 1;
char s[N];
void work(int s, int t) {
  if (s == 2) f[2] += t;
  if (s == 3) f[3] += t;
  if (s == 4) f[2] += 2 * t;
  if (s == 5) f[5] += t;
  if (s == 6) {
    f[2] += t;
    f[3] += t;
  }
  if (s == 7) f[7] += t;
  if (s == 8) f[2] += 3 * t;
  if (s == 9) f[3] += 2 * t;
  return;
}
int main() {
  scanf("%d%s", &n, s);
  for (int i = 0; i < n; i++)
    for (int j = 2; j <= s[i] - '0'; j++) a[j]++;
  for (int i = 2; i < 10; i++)
    if (a[i]) ok = 0;
  if (ok) {
    puts("0");
    return 0;
  }
  for (int i = 2; i < 10; i++)
    while (a[i]--) work(i, 1);
  for (int i = 9; i > 1; i--)
    while (f[i]) {
      putchar(i + '0');
      for (int j = i; j > 1; j--) work(j, -1);
    }
  return 0;
}

```