#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e6 + 3;
const double PI = acos(-1);
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y == 1)
    return x;
  long long r = power(x, y / 2);
  if (y % 2 == 0)
    return (r * r);
  else
    return (r * r * x);
}
vector<bool> prime(10000001, true);
void sieveprime() {
  prime[0] = false;
  prime[1] = false;
  for (int i = 2; i < prime.size(); i++) {
    if (prime[i] == true) {
      for (int j = i + i; j <= 10000000; j += i) prime[j] = false;
    }
  }
  return;
}
long long binarytodec(string d) {
  long long s = 1;
  long long ans = 0;
  while (d.size() > 0) {
    char lst = d.back();
    int ld = lst - '0';
    ans += (ld * s) % mod;
    s = (s % mod) * 2;
    d.pop_back();
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  string s;
  cin >> s;
  string x = "";
  map<char, string> mp;
  mp['2'] = "2";
  mp['3'] = "3";
  mp['4'] = "322";
  mp['5'] = "5";
  mp['6'] = "53";
  mp['7'] = "7";
  mp['8'] = "7222";
  mp['9'] = "7332";
  for (int i = 0; i < n; i++) {
    if (s[i] != '1' && s[i] != '0') x += mp[s[i]];
  }
  sort(x.rbegin(), x.rend());
  cout << x;
  return 0;
}

### 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 long long mod = 1e6 + 3;
const double PI = acos(-1);
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y == 1)
    return x;
  long long r = power(x, y / 2);
  if (y % 2 == 0)
    return (r * r);
  else
    return (r * r * x);
}
vector<bool> prime(10000001, true);
void sieveprime() {
  prime[0] = false;
  prime[1] = false;
  for (int i = 2; i < prime.size(); i++) {
    if (prime[i] == true) {
      for (int j = i + i; j <= 10000000; j += i) prime[j] = false;
    }
  }
  return;
}
long long binarytodec(string d) {
  long long s = 1;
  long long ans = 0;
  while (d.size() > 0) {
    char lst = d.back();
    int ld = lst - '0';
    ans += (ld * s) % mod;
    s = (s % mod) * 2;
    d.pop_back();
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  string s;
  cin >> s;
  string x = "";
  map<char, string> mp;
  mp['2'] = "2";
  mp['3'] = "3";
  mp['4'] = "322";
  mp['5'] = "5";
  mp['6'] = "53";
  mp['7'] = "7";
  mp['8'] = "7222";
  mp['9'] = "7332";
  for (int i = 0; i < n; i++) {
    if (s[i] != '1' && s[i] != '0') x += mp[s[i]];
  }
  sort(x.rbegin(), x.rend());
  cout << x;
  return 0;
}

```