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 = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, true) : false;
}
int fact[10];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  string str;
  cin >> str;
  for (int i = 0; i < (int)(str.size()); i++) {
    if (str[i] == '2') fact[2]++;
    if (str[i] == '3') fact[3]++;
    if (str[i] == '4') {
      fact[2] += 2;
      fact[3]++;
    }
    if (str[i] == '5') fact[5]++;
    if (str[i] == '6') {
      fact[5]++;
      fact[3]++;
    }
    if (str[i] == '7') fact[7]++;
    if (str[i] == '8') {
      fact[7]++;
      fact[2] += 3;
    }
    if (str[i] == '9') {
      fact[7]++;
      fact[2] += 1;
      fact[3] += 2;
    }
  }
  for (int i = 9; i >= (int)(0); i--)
    for (int j = 0; j < (int)(fact[i]); j++) cout << i;
}
