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 <algorithm>
#include <bitset>
#include <cmath>
#include <functional>
#include <iostream>
#include <iterator>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <utility>
#include <unordered_map>
#include <unordered_set>
using namespace std;
 
#define ABS(a) (((a) >= 0) ? (a) : -(a))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) > (b) ? (b) : (a))
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define RFOR(i, a, b) for (int i = (a); i >= (b); --i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define POW_10_9 1000000000
#define INF (numeric_limits<int>::max()/2)
#define int long long
 
typedef long long ll;
typedef pair<int, int> Pair;

const int costs[10] = {0,2,5,5,4,5,6,3,7,6};
int N,M;
string ans;
int A[9];
//桁DP
int dp[10001];

signed main(){
    cin >> N >> M;
    REP(i,M){cin>>A[i];}
    sort(A,A+M,greater<int>());//降順にソート
    FOR(i,1,N){
        REP(j,M){
            // 桁DP
            int k = i-costs[A[j]];
            if(k>=0) dp[i] = MAX(dp[i],dp[k]+1);
        }
        if(dp[i] == 0) dp[i] = -INF; //一つの整数のみで達成される場合
    }
    int i=N;
    REP(j,M){
        int k = i-costs[A[j]];
        while(k>=0 && dp[i]==dp[k]+1){
            ans.push_back('0'+A[j]);
            i -= costs[A[j]];
            k = i-costs[A[j]];
        }
    }
    cout << ans << endl;

    return 0;
}