Problem:
Score : 400 points

Problem Statement

Find the largest integer that can be formed with exactly N matchsticks, under the following conditions:

Every digit in the integer must be one of the digits A_1, A_2, ..., A_M (1 \leq A_i \leq 9).

The number of matchsticks used to form digits 1, 2, 3, 4, 5, 6, 7, 8, 9 should be 2, 5, 5, 4, 5, 6, 3, 7, 6, respectively.

Constraints

All values in input are integers.

2 \leq N \leq 10^4

1 \leq M \leq 9

1 \leq A_i \leq 9

A_i are all different.

There exists an integer that can be formed by exactly N matchsticks under the conditions.

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_M

Output

Print the largest integer that can be formed with exactly N matchsticks under the conditions in the problem statement.

Sample Input 1

20 4
3 7 8 4

Sample Output 1

777773

The integer 777773 can be formed with 3 + 3 + 3 + 3 + 3 + 5 = 20 matchsticks, and this is the largest integer that can be formed by 20 matchsticks under the conditions.

Sample Input 2

101 9
9 8 7 6 5 4 3 2 1

Sample Output 2

71111111111111111111111111111111111111111111111111

The output may not fit into a 64-bit integer type.

Sample Input 3

15 3
5 4 6

Sample Output 3

654

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=1e9+7;

int c[10]={0,2,5,5,4,5,6,3,7,6};

int main(){
  int n,m;
  cin>>n>>m;
  int a[m];
  rep(i,0,m) cin>>a[i];
  sort(a,a+m);
  reverse(a,a+m);
  int dp[n+1];
  rep(i,0,n+1) dp[i]=0;
  rep(i,1,n+1){
    rep(j,0,m){
      if(i>=c[a[j]]){
        dp[i]=max(dp[i],dp[i-c[a[j]]]+1);
      }
    }
  }
  //rep(i,0,n+1) cout<<dp[i];
  int cur=n;
  vector<int> v;
  while(cur>0){
    rep(i,0,m){
      if(cur-c[a[i]]>=0 && dp[cur]==dp[cur-c[a[i]]]+1){
        if(dp[cur]!=1){
          cur-=c[a[i]];
          v.push_back(a[i]);
          break;
        }
        else{
          if(cur==c[a[i]]){
            cur-=c[a[i]];
            v.push_back(a[i]);
            break;
          }
        }
      }
    }
  }
  for(int p:v) cout<<p;
  cout<<endl;
}