#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
long long C(int n, int r) {
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  long long ans = 1;
  int i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
vector<long long> sieve(long long n) {
  vector<long long> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
      for (int j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
  return is_prime;
}
void test() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> freq(10);
  for (auto &x : s) {
    int c = x - '0';
    if (c == 0 || c == 1)
      continue;
    else if (c == 2)
      ++freq[2];
    else if (c == 3)
      freq[3]++;
    else if (c == 4)
      freq[2] += 2, freq[3]++;
    else if (c == 5)
      freq[5]++;
    else if (c == 6)
      freq[3]++, freq[5]++;
    else if (c == 7)
      freq[7]++;
    else if (c == 8)
      freq[7]++, freq[2] += 3;
    else if (c == 9)
      freq[7]++, freq[2]++, freq[3] += 2;
  }
  string ans = "";
  for (int i = 7; i >= 2; --i) {
    for (int k = 0; k < freq[i]; ++k) ans.push_back(i + '0');
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long T = 0; T < t; T++) {
    test();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

### Prompt

Generate 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;
int mod = 1e9 + 7;
long long C(int n, int r) {
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  long long ans = 1;
  int i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
vector<long long> sieve(long long n) {
  vector<long long> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
      for (int j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
  return is_prime;
}
void test() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> freq(10);
  for (auto &x : s) {
    int c = x - '0';
    if (c == 0 || c == 1)
      continue;
    else if (c == 2)
      ++freq[2];
    else if (c == 3)
      freq[3]++;
    else if (c == 4)
      freq[2] += 2, freq[3]++;
    else if (c == 5)
      freq[5]++;
    else if (c == 6)
      freq[3]++, freq[5]++;
    else if (c == 7)
      freq[7]++;
    else if (c == 8)
      freq[7]++, freq[2] += 3;
    else if (c == 9)
      freq[7]++, freq[2]++, freq[3] += 2;
  }
  string ans = "";
  for (int i = 7; i >= 2; --i) {
    for (int k = 0; k < freq[i]; ++k) ans.push_back(i + '0');
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long T = 0; T < t; T++) {
    test();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

```