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;
typedef long long ll;
#define rep(i, j, n) for (ll i = j; i < n; i++)
#define all(x) (x).begin(),(x).end()
#define INF (1000000000)
#define MOD (1000000007)
#define MAX (110)
#define pii pair<int, int>

/////////////////////////////////////////////////////////

bool comp(pii a, pii b){
  if(a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}

void Main() {
  vector<int> list = {INF, 2, 5, 5, 4, 5, 6, 3, 7, 6};
  int N, M; cin >> N >> M;
  vector<pii> A(M);
  rep(i, 0, M){
    int tmp; cin >> tmp;
    A[i] = make_pair(tmp, list[tmp]);
  }
  sort(all(A), comp);

  vector<int> dp(N + 1, -INF);
  dp[0] = 0;
  rep(i, 1, N + 1){
    rep(j, 0, M){
      if(i >= A[j].second) if(dp[i] < dp[i - A[j].second] + 1) dp[i] = dp[i - A[j].second] + 1;
    }
  }

  int num = dp[N];
  vector<int> res(num);
  int n = N;
  rep(i, 0, num){
    rep(j, 0, M){
      if(n < A[j].second) continue;
      if(dp[n] == dp[n - A[j].second] + 1){
        res[i] = A[j].first;
        n -= A[j].second;
        break;
      }
    }
  }
  
  rep(i, 0, num) cout << res[i];
  cout << endl;
}

/////////////////////////////////////////////////////////

int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  //cout << std::fixed << std::setprecision(15);

  Main();
}