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;
int num[10][5];
int cnt[10][5];
int n;
string str;
int sum[10];
signed main() {
  num[2][1] = 1;
  num[3][2] = 1;
  num[4][1] = 2;
  num[5][3] = 1;
  num[6][1] = 1;
  num[6][2] = 1;
  num[7][4] = 1;
  num[8][1] = 3;
  num[9][2] = 2;
  for (int i = 2; i <= 9; i++)
    for (int j = 1; j <= 4; j++) cnt[i][j] = cnt[i - 1][j] + num[i][j];
  cin >> n;
  for (int i = 1; i <= 4; i++) sum[i] = 0;
  cin >> str;
  int len = str.length();
  for (int i = 0; i < len; i++) {
    int t = str[i] - '0';
    for (int j = 1; j <= 4; j++) sum[j] += cnt[t][j];
  }
  for (int i = 1; i <= sum[4]; i++)
    cout << 7, sum[3] -= cnt[7][3], sum[2] -= cnt[7][2], sum[1] -= cnt[7][1];
  for (int i = 1; i <= sum[3]; i++)
    cout << 5, sum[2] -= cnt[5][2], sum[1] -= cnt[5][1];
  for (int i = 1; i <= sum[2]; i++) cout << 3, sum[1] -= cnt[3][1];
  for (int i = 1; i <= sum[1]; i++) cout << 2;
}
