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>
#include <bits/stdc++.h>
using namespace std;
const int inf = 1000000009;
int rd() {
  char c = getchar();
  int t = 0, f = 1;
  while (!isdigit(c)) f = (c == '-') ? -1 : 1, c = getchar();
  while (isdigit(c)) t = t * 10 + c - 48, c = getchar();
  return t * f;
}
void wt(int x) {
  if (x > 9) wt(x / 10);
  putchar(x % 10 + 48);
  return;
}
int n;
char str[50];
int num[20];
int main() {
  n = rd();
  scanf("%s", str + 1);
  for (int i = ((int)1); i <= ((int)n); i++) num[str[i] - 48]++;
  num[3] += num[4], num[2] += num[4] * 2;
  num[4] = 0;
  num[5] += num[6], num[3] += num[6];
  num[6] = 0;
  num[2] += 3 * num[8], num[7] += num[8];
  num[8] = 0;
  num[2] += num[9], num[3] += num[9] * 2, num[7] += num[9];
  for (int i = ((int)8); i >= ((int)2); i--)
    while (num[i]--) putchar(i + 48);
  return 0;
  return 0;
}
