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 <bits/stdc++.h>

using namespace std;


//repetition
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define euc_dis(x, y) sqrt(x*x + y*y)

#define MOD 1e9+7
#define sp ' '


using ll = long long;


ll mod(ll a, ll b){return (a%b+b)%b;}
ll gcd(ll a, ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
void Yes(){cout << "Yes" << endl;}
void No(){cout << "No" << endl;}
void Judge(bool b){b?Yes():No();}
void YES(){cout << "YES" << endl;}
void NO(){cout << "NO" << endl;}
void JUDGE(bool b){b?YES():NO();}
ll powMod(ll b, ll e, ll m){ll r=1;while(e>0){if(e&1)r=(r%m)*(b%m)%m;b=(b%m)*(b%m)%m;e>>=1;}return r;}
double distance(ll x1, ll y1, ll x2, ll y2){return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}

template<typename T>
void ppp(T n){cout << n << endl;}

string nchar(char c, int n){
    string s = "";
    rep(i, n){
        s.push_back(c);
    }

    return s;
}

int main(){

    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, m;
    cin >> n >> m;

    string ans = "";
    vector<pair<int, int>> digit(10);
    vector<int> a(n);
    int num[9] = {2,5,5,4,5,6,3,7,6};
    rep(i, 9){
        digit[i+1] = make_pair(num[i], i+1);
    }

    int sum = 0;
    rep(i, m){
        cin >> a[i];
        sum += a[i];
    }

    vector<pair<int,int>> use(m);
    rep(i, m){
        use[i] = digit[a[i]];
    }
    sort(use.begin(), use.end());

    int remain = n;
    int idx = 0;

    if(sum == n){
        for(int i = 0; i < m; ++i){
            ans += a[i]+'0';
        }
        sort(ans.rbegin(), ans.rend());
        ppp(ans);
        return 0;
    }
    while(remain > 0){
        if(idx >= m) break;
        auto mn = use[idx];
        if(mn.first == 5 and mn.second == 2){
            mn.first = 5;
            mn.second = 5;
        }
        if(mn.first == 5 and mn.second == 5){
            mn.first = 5;
            mn.second = 2;
        }
        if(mn.first == 6 and mn.second == 6){
            mn.first = 6;
            mn.second = 9;
        }
        if(mn.first == 6 and mn.second == 9){
            mn.first = 6;
            mn.second = 6;
        }
        if(remain - mn.first >= mn.first or remain == mn.first){
            ans += mn.second+'0';
            remain -= mn.first;
        }
        else{
            idx++;
        }
    }

    sort(ans.rbegin(), ans.rend());
    ppp(ans);
    return 0;
}