Drazil is playing a math game with Varda.

Let's define <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7020b94c48be2ee5b8eeca05f379c8911bb0e7c4.png" style="max-width: 100.0%;max-height: 100.0%;"/> for positive integer *x* as a product of factorials of its digits. For example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6784d861d9a724d0aeddefcb00b7d0c3442ae586.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

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. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7020b94c48be2ee5b8eeca05f379c8911bb0e7c4.png" style="max-width: 100.0%;max-height: 100.0%;"/> = <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/aa57e017a6c74d06b0a54bc47e3f6990e5af38c1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

Help friends find such number.
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.
#include <bits/stdc++.h>
#define ll long long
#define sp ' '
//#define endl "\n"
#define pb push_back
#define ar for(int i=1;i<=n;i++)
using namespace std;
ll t=1;
const ll N=5e5+100;
ll n,m,k,b[N],a[N];
ll ans=0;
ll p[11][N];
void solve()
{
    map<int,int>mp;
    cin>>n;
    vector<char>v;
    for(int i=1;i<=n;i++){char x;cin>>x;v.push_back(x);}
    for(auto i:v)
    {

        ll x=i-'0';
        if(x==2){mp[2]++;}
        else if(x==3){mp[3]++;}
        else if(x==4){mp[2]+=2;mp[3]++;}
        else if(x==5){mp[5]++;}
        else if(x==6){mp[5]++;mp[3]++;}
        else if(x==7){mp[7]++;}
        else if(x==8){mp[7]++;mp[2]+=3;}
        else if(x==9){mp[7]++;mp[2]++;mp[3]+=2;}
    }
    for(int i=9;i>=2;i--)
    {
        for(int j=1;j<=mp[i];j++)
        {
            cout<<i;
        }
    }
    return;
}

int main()
{
    ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
    //cin>>t;
    while(t--){solve();}return 0;
}
