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: Accepted)

/*|In The Name Of Allah|*/

#include <bits/stdc++.h>

using namespace std;

const int N = 1e4+5;

int n;
int m;
int valid;
int mem[N];
int cost[10];

int go(int lft){
    if(lft == 0)
        return 0;
    if(~mem[lft])
        return mem[lft];
    int bst = -1e2;
    for(int i = 1; i <= 9; i++){
        if(valid & (1 << i) && lft >= cost[i]){
            int ret = go(lft - cost[i]);
            if(ret != -1e2)
                bst = max(bst , ret + 1);
        }
    }
    return mem[lft] = bst;
}

vector < int > mine;

void build(int lft){
    for(int i = 9; i >= 1; i--){
        if(valid & (1 << i) && lft >= cost[i] && go(lft) >= 0 && go(lft - cost[i]) >= 0 && go(lft) == go(lft - cost[i]) + 1){
            mine.push_back(i);
            build(lft - cost[i]);
            break;
        }
    }
}

int main(){
    memset(mem , -1 , sizeof mem);
    cost[1] = 2;
    cost[2] = 5;
    cost[3] = 5;
    cost[4] = 4;
    cost[5] = 5;
    cost[6] = 6;
    cost[7] = 3;
    cost[8] = 7;
    cost[9] = 6;
    scanf("%d%d" , &n , &m);
    for(int i = 1; i <= m; i++){
        int x;
        scanf("%d" , &x);
        valid |= 1 << x;
    }
    int sz = go(n);
    build(n);
    sort(mine.rbegin() , mine.rend());
    for(auto v : mine)
        printf("%d" , v);
}