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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

const
	t:array[1..9] of longint=(2,5,5,4,5,6,3,7,6);
var
	n,m,t1,i,j,cnt:longint;
	a,ans:array[-1100..10010] of longint;
	f:array[-1100..10010] of boolean;
procedure qsort(l,r:longint);
var
	i,j,t,mid:longint;
begin
	i:=l; j:=r; mid:=ans[(l+r) div 2];
	repeat
		while ans[i]<mid do inc(i);
		while ans[j]>mid do dec(j);
		if i<=j then
		begin
			t:=ans[i]; ans[i]:=ans[j]; ans[j]:=t;
			inc(i); dec(j);
		end;
	until i>j;
	if l<j then qsort(l,j);
	if i<r then qsort(i,r);
end;
begin
	readln(n,m);
	fillchar(f,sizeof(f),false);
	f[0]:=true;
	for i:=1 to m do
	begin
		read(a[i]);
		for j:=t[a[i]] to n do
			if f[j-t[a[i]]] then f[j]:=true;
	end;
	cnt:=0;
	for i:=1 to m-1 do
		for j:=1 to m-i do
			if (t[a[j]]>t[a[j+1]]) or (t[a[j]]=t[a[j+1]]) and (a[j]<a[j+1]) then
			begin
				t1:=a[j]; a[j]:=a[j+1]; a[j+1]:=t1;
			end;
	t1:=1;
	while t1<=m do
	begin
		while f[n-t[a[t1]]] and (n-t[a[t1]]>=0) do
		begin
			n:=n-t[a[t1]];
			inc(cnt); ans[cnt]:=a[t1];
		end;
		inc(t1);
	end;
	qsort(1,cnt);
	for i:=cnt downto 1 do write(ans[i]);
	writeln;
end.