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;

/* macro */
#define rep(i,a,b) for(int i=a;i<b;i++)
#define revrep(i,a,b) for(int i = a; i > b; i--)
#define int long long
#define exist(s,e) ((s).find(e)!=(s).end())
#define all(v) (v).begin(), (v).end()
#define each(s,itr) for(auto (itr) = s.begin(); (itr) != s.end(); (itr)++)
#define sum(v) accumulate(all(v), (0LL))
#define isin(a, b, c) (b <= a && a <= c)
#define println cout << "\n";
#define sz(v) (int)v.size()


/* alias */
template<class T> using vec = vector<T>;
typedef vector<int> vi;
typedef pair<int, int> pi;
typedef tuple<int, int, int> ti;
typedef map<int, int> mi;
typedef set<int> si;


/* constant */
const int inf = 1LL << 62;
const int mod = 1e9 + 7;
const int dx[8]={1,0,-1,0,-1,1,-1,1};
const int dy[8]={0,1,0,-1,-1,-1,1,1};
const string alphabet = "abcdefghijklmnopqrstuvwxyz";

/* io_method */
int input(){int tmp;cin >> tmp;return tmp;}
string raw_input(){string tmp;cin >> tmp;return tmp;}
template<class T> void printx(T n){cout << n;}
template<class T, class U> void printx(pair<T, U> p){cout << "(" << p.first << "," << p.second << ")";}
template<class T, class U, class V> void printx(tuple<T, U, V> t){cout << "{" << get<0>(t) << "," << get<1>(t) <<"," << get<2>(t) << "}" << endl;}
template<class T> void printx(vector<T> v){cout << "{";rep(i,0,v.size()){printx(v[i]);if(i != v.size()-1)printx(",");}cout << "}";}
template<class T> void print(T n){printx(n);cout << endl;}
template<class T> void print(set<T> s){cout << "{";each(s, e){if(e != s.begin()) printx(",");printx(*e);}cout << "}" << endl;}
template<class T, class U> void print(map<T, U> mp){cout << "{";each(mp, e){cout << "(" << e -> first << "," << e -> second << ")";}cout << "}" << endl;}
template<class T> void printans(vec<T> v){rep(i,0,sz(v)){cout << v[i] << (i == sz(v) - 1 ? "" : " ");}cout << endl;}


/* general_method */
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T>T cut(T &a, int l, int r){return T(a.begin()+l, a.begin()+r);}

/* main */

int n, m;
si arr;
int need[10] = {-1, 2, 5, 5, 4, 5, 6, 3, 7, 6};
void in(){
    cin >> n >> m;
    rep(i,0,m) arr.insert(input());
}

bool isbig(string a, string b){
    if(sz(a) < sz(b)) return true;
    else if(sz(a) == sz(b)){
        rep(i,0,sz(a)){
            if(a[i] > b[i]) return false;
            if(a[i] < b[i]) return true;
        }
        return false;
    }else{
        return false;
    }
}

string dp[10010];

string dfs(int rest){


    if(rest == 0) return "&";
    if(rest < 0) return "$";

    if(dp[rest] != "-1") return dp[rest];
    string ans = "";
    rep(i,1,10){
        if(exist(arr, i)){
            string tmp = dfs(rest - need[i]);
            if(tmp.back() == '$') continue;
            if(isbig(ans, to_string(i) + tmp)){
                ans = to_string(i) + tmp;
            }
        }
    }

    return dp[rest] = ans;


}


void solve(){

    rep(i,0,10010) dp[i] = "-1";
    string ans = dfs(n);
    print(cut(ans, 0, sz(ans) - 1));



}


signed main(){

    cin.tie(0);
    ios::sync_with_stdio(false);
    in();
    solve();
    
}