#include <bits/stdc++.h>
using namespace std;
const long long int MAX = 9187201950435737471;
const double pi = acos(-1);
const int N = 129;
const long long mod = 1000000007;
const int M = 2147483647;
const int lg = 101;
int n;
long long int A[N];
long long int child[N];
set<long long int> s;
map<long long int, long long int> mp;
bool dfs(long long int x, int cnt) {
  if (cnt == n) return true;
  bool ret = 0;
  if (x % 3 == 0) {
    if (s.count(x / 3)) {
      child[x] = mp[x / 3];
      ret |= dfs(x / 3, cnt + 1);
    }
  }
  if (s.count(x * 2)) {
    child[x] = mp[x * 2];
    ret |= dfs(x * 2, cnt + 1);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (s[i] == '4') {
      v.push_back(2);
      v.push_back(2);
      v.push_back(3);
    } else if (s[i] == '8') {
      v.push_back(7);
      v.push_back(2);
      v.push_back(2);
      v.push_back(2);
    } else if (s[i] == '9') {
      v.push_back(7);
      v.push_back(3);
      v.push_back(3);
      v.push_back(2);
    } else if (s[i] == '6') {
      v.push_back(5);
      v.push_back(3);
    } else {
      if (s[i] - '1' > 0) v.push_back(s[i] - '0');
    }
  }
  sort(v.begin(), v.end());
  int m = (int)v.size();
  ;
  for (int i = m - 1; i >= 0; i--) cout << v[i];
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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>
using namespace std;
const long long int MAX = 9187201950435737471;
const double pi = acos(-1);
const int N = 129;
const long long mod = 1000000007;
const int M = 2147483647;
const int lg = 101;
int n;
long long int A[N];
long long int child[N];
set<long long int> s;
map<long long int, long long int> mp;
bool dfs(long long int x, int cnt) {
  if (cnt == n) return true;
  bool ret = 0;
  if (x % 3 == 0) {
    if (s.count(x / 3)) {
      child[x] = mp[x / 3];
      ret |= dfs(x / 3, cnt + 1);
    }
  }
  if (s.count(x * 2)) {
    child[x] = mp[x * 2];
    ret |= dfs(x * 2, cnt + 1);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (s[i] == '4') {
      v.push_back(2);
      v.push_back(2);
      v.push_back(3);
    } else if (s[i] == '8') {
      v.push_back(7);
      v.push_back(2);
      v.push_back(2);
      v.push_back(2);
    } else if (s[i] == '9') {
      v.push_back(7);
      v.push_back(3);
      v.push_back(3);
      v.push_back(2);
    } else if (s[i] == '6') {
      v.push_back(5);
      v.push_back(3);
    } else {
      if (s[i] - '1' > 0) v.push_back(s[i] - '0');
    }
  }
  sort(v.begin(), v.end());
  int m = (int)v.size();
  ;
  for (int i = m - 1; i >= 0; i--) cout << v[i];
  return 0;
}

```