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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
#define ll long long
using namespace std;
int main()
{
    vector<int> dp(10000);
    int n, m;
    cin >> n >> m;
    vector<int> a(n),use={0,2,5,5,4,5,6,3,7,6};
    rep(i,m){
        cin >> a[i];
    }
    for(auto e:a){
        dp[use[e]]++;
    }
    rep(i,n+1){
        if(dp[i]==0)
            continue;
        for(auto e:a){
            dp[i + use[e]] = max(dp[i] + 1, dp[i + use[e]]);
        }
    }
    string ans;
    int remain = dp[n];
    int match = n;
    sort(a.rbegin(), a.rend());
    while(match > 0){
        for(auto e: a){
            //ここから遷移してきたものが最大桁になるとわかる
            //しかもaを並べ替えておいたのでおっきい順に試してくれる
            if(dp[match-use[e]]==remain-1){
                ans += char(e+'0');
                match -= use[e];
                remain--;
                break;
            }
        }        
    }
    cout << ans << endl;
}