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 double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 2000 * 1000 * 1000;
template <typename T>
inline T sqr(T n) {
  return n * n;
}
vector<int> mapping[10] = {{},  {},     {2}, {3},          {3, 2, 2},
                           {5}, {5, 3}, {7}, {7, 2, 2, 2}, {7, 3, 3, 2}};
int n;
char s[30];
vector<int> result;
int main() {
  scanf("%d\n", &n);
  gets(s);
  for (int i = 0; i < n; i++) {
    for (size_t j = 0; j < mapping[s[i] - '0'].size(); j++) {
      result.push_back(mapping[s[i] - '0'][j]);
    }
  }
  sort((result).begin(), (result).end());
  reverse((result).begin(), (result).end());
  for (size_t i = 0; i < result.size(); i++) {
    printf("%d", result[i]);
  }
  puts("");
  return 0;
}
