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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define ln '\n'
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
#define FOR(i, a, b) for(int i = (a); i <= (int)(b); ++i)
#define FORR(i, a, b) for(int i = (a); i >= (int)(b); --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
typedef vector<P> VP;
typedef vector<PL> VPL;
typedef vector<string> VS;

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


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    VI a(m);
    REP(i, m) cin >> a[i];
    vector<string> dp(n + 10, "");

    REP(i, m) dp[stick[a[i]]] = (char)('0' + a[i]);
    REP(i, n + 1) {
        string max_str = "";
        REP(j, m) {
            if(i - stick[a[j]] < 0) continue;
            if(dp[i - stick[a[j]]] == "") continue;
            string cand = dp[i - stick[a[j]]] + (char)('0' + a[j]);
            
            if(max_str.size() < cand.size()) 
                max_str = cand;
            else if(max_str.size() == cand.size() and max_str < cand)
                max_str = cand;
        }

        if(dp[i].size() < max_str.size()) 
            dp[i] = max_str;
        else if(dp[i].size() == max_str.size() and dp[i] < max_str)
            dp[i] = max_str;
    }

    // REP(i, n + 1) cout << i << " " << dp[i] << ln;
    cout << dp[n] << ln;
    return 0;
}