#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
vector<int> v[100005];
int vis[100005];
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
int binomialCoeff(int n, int k) {
  if (k == 0 || k == n) return 1;
  return binomialCoeff(n - 1, k - 1) + binomialCoeff(n - 1, k);
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % MOD;
    b /= 2;
    a = (a * a) % MOD;
  }
  return ans;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int sieve(int n) {
  if (n <= 1) {
    return 0;
  }
  if (n <= 3) {
    return 1;
  }
  if ((n % 2 == 0) || (n % 3 == 0)) {
    return 0;
  }
  int k = 5;
  while (k * k <= n) {
    if ((n % k == 0) || (n % (k + 2) == 0)) {
      return 0;
    }
    k = k + 6;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (s[i] == '2') v.push_back(2);
    if (s[i] == '3') v.push_back(3);
    if (s[i] == '4') {
      v.push_back(2);
      v.push_back(2);
      v.push_back(3);
    }
    if (s[i] == '5') v.push_back(5);
    if (s[i] == '6') {
      v.push_back(3);
      v.push_back(5);
    }
    if (s[i] == '7') v.push_back(7);
    if (s[i] == '8') {
      v.push_back(2);
      v.push_back(2);
      v.push_back(2);
      v.push_back(7);
    }
    if (s[i] == '9') {
      v.push_back(2);
      v.push_back(3);
      v.push_back(3);
      v.push_back(7);
    }
  }
  sort(v.begin(), v.end());
  for (int i = v.size() - 1; i >= 0; i--) {
    cout << v[i];
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 int MOD = 1e9 + 7;
vector<int> v[100005];
int vis[100005];
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
int binomialCoeff(int n, int k) {
  if (k == 0 || k == n) return 1;
  return binomialCoeff(n - 1, k - 1) + binomialCoeff(n - 1, k);
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % MOD;
    b /= 2;
    a = (a * a) % MOD;
  }
  return ans;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int sieve(int n) {
  if (n <= 1) {
    return 0;
  }
  if (n <= 3) {
    return 1;
  }
  if ((n % 2 == 0) || (n % 3 == 0)) {
    return 0;
  }
  int k = 5;
  while (k * k <= n) {
    if ((n % k == 0) || (n % (k + 2) == 0)) {
      return 0;
    }
    k = k + 6;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (s[i] == '2') v.push_back(2);
    if (s[i] == '3') v.push_back(3);
    if (s[i] == '4') {
      v.push_back(2);
      v.push_back(2);
      v.push_back(3);
    }
    if (s[i] == '5') v.push_back(5);
    if (s[i] == '6') {
      v.push_back(3);
      v.push_back(5);
    }
    if (s[i] == '7') v.push_back(7);
    if (s[i] == '8') {
      v.push_back(2);
      v.push_back(2);
      v.push_back(2);
      v.push_back(7);
    }
    if (s[i] == '9') {
      v.push_back(2);
      v.push_back(3);
      v.push_back(3);
      v.push_back(7);
    }
  }
  sort(v.begin(), v.end());
  for (int i = v.size() - 1; i >= 0; i--) {
    cout << v[i];
  }
}

```