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 INF (INT32_MAX)
#define MOD (1000000000+7)
#define MAX 9

int N, M;
int A[MAX];

int num[9] = { 2, 5, 5, 4, 5, 6, 3, 7, 6 };
int dp[10000 + 1];

int main() {
	cin >> N >> M;
	for( int i = 0; i < M; i++ ) {
		cin >> A[i];
	}

	for( int i = 0; i <= N; i++ ) {
		dp[i] = -INF;
	}
	dp[0] = 0;
	for( int i = 1; i <= N; i++ ) {
		for( int j = 0; j < M; j++ ) {
			int a = A[j] - 1;
			int n = num[a];
			if( i - n >= 0 ) {
				dp[i] = max( dp[i], dp[i - n] + 1 );
			}
		}
	}
	int d = dp[N];
	sort( A, A + M, greater<int>() );
	string ans;
	for( int i = 0; i < d; i++ ) {
		for( int j = 0; j < M; j++ ) {
			int a = A[j] - 1;
			int n = num[a];
			if( N - n >= 0 && dp[N - n] == dp[N] - 1 ) {
				ans += (char)(A[j] + '0');
				N -= n;
				break;
			}
		}
	}

	cout << ans << endl;

	return 0;
}