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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
#include <complex>
#include <stack>
#include <queue>
#include <unordered_map>
#include <map>
#define rep(i, a) for (int i = 0; i < (a); i++)
using namespace std;

int main(){
    int dp[10001] = {-10000};
    vector<int> v;
    int i, j, n, m, a;
    cin >> n >> m;
    
    int num[9] = {2, 5, 5, 4, 5, 6, 3, 7, 6};
    
    for(i = 0; i < m; i++){
        cin >> a;
        v.push_back(a);
    }
    sort(v.begin(), v.end());
    
    dp[0] = 0;
    dp[1] = 0;
    
    for(i = 2; i <= n; i++){
        for(j = 0; j < m; j++){
            if(i >= num[v[j]-1]){
                dp[i] = max(dp[i], dp[i-num[v[j]-1]]+1);
            }
        }
    }
    
    /*
    for(i = 0; i <= n; i++){
        cout << dp[i] << endl;
    }
     */
    
    int N = n;
    int flag = 0;
    while(1){
        for(i = m-1; i >= 0; i--){
            if(dp[N-num[v[i]-1]] == dp[N]-1){
                if(dp[N]-1 == 0){
                    if(N-num[v[i]-1] == 0){
                        flag = 1;
                    }else{
                        continue;
                    }
                }
                N -= num[v[i]-1];
                cout << v[i];
                break;
            }
        }
        if(flag){
            break;
        }
    }
    
    cout << endl;
}