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)

#include<bits/stdc++.h>

using namespace std;

#define all(x) x.begin(), x.end()
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef long long ll;
typedef long double ld;

const int mod = 1e9 + 7;
const ll inf = 3e18 + 5;

const int maxn = 1e4 + 5;
int a[maxn], dp[maxn], m, nxt[maxn];
map <int, int> mp;

int f(int n){
    if(n < 0)
        return -mod;
    if(n == 0)
        return 0;
    if(dp[n] != -1)
        return dp[n];
    int ret = -1;
    for(int i = m - 1; i >= 0; i--){
        if(1 + f(n - mp[a[i]]) > ret){
            nxt[n] = a[i];
            ret = 1 + f(n - mp[a[i]]);
        }
    }
    return dp[n] = ret;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n >> m;
    int h[] = {2, 5, 5, 4, 5, 6, 3, 7, 6};
    for(int i = 1; i <= 9; i++)
        mp[i] = h[i - 1];
    for(int i = 0; i < m; i++){
        cin >> a[i];
    }
    memset(dp, -1, sizeof dp);
    sort(a, a + m);
    f(n);
    while(n > 0){
        cout << nxt[n];
        n -= mp[nxt[n]];
    }
}