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<iostream>
#include<vector>
#include<string>
#include<algorithm>

using namespace std;

const string x = "-";

void chmax(string &a, string b){
  if ( a == x){ a = b;}
  else if( a.size() < b.size()){ a = b;}
  else if( a == b){
    a  = max( a , b);}

}

int match(int a){
  if( a == 1){ return 2;}
  else if( a == 2){ return 5;}
  else if( a == 3){ return 5;}
  else if( a == 4){ return 4;}
  else if( a == 5){ return 5;}
  else if( a == 6){ return 6;}
  else if( a == 7){ return 3;}
  else if( a == 8){ return 7;}
  else if( a == 9){ return 6;}
}
                   

int main(){
  int N,M; cin >> N >> M;
  vector<int> A(M);
  for(int i = 0; i < M; i++){ cin >> A[i];}
  
  string dp[10100];
  
  for(int i  =0; i < 10100; i++){ dp[i] = x;}
  
  dp[0] = ""; //配るDP "-"は未更新 ""は更新したが何も書けない
  
  for(int i = 0; i < 10100; i++){
    
   if( dp[i] == x){continue;} //未更新から配ってもしょうがない
   
    for(int j = 0; j < M; j++){
      chmax( dp[i+match(A[j])], dp[i] + to_string(A[j]));}
  }
                                
                                cout << dp[N] << endl; return 0;}