Problem: You are given a set S of strings consisting of `0` and `1`, and an integer K.

Find the longest string that is a subsequence of K or more different strings in S. If there are multiple strings that satisfy this condition, find the lexicographically smallest such string.

Here, S is given in the format below:

* The data directly given to you is an integer N, and N+1 strings X_0,X_1,...,X_N. For every i (0\leq i\leq N), the length of X_i is 2^i.
* For every pair of two integers (i,j) (0\leq i\leq N,0\leq j\leq 2^i-1), the j-th character of X_i is `1` if and only if the binary representation of j with i digits (possibly with leading zeros) belongs to S. Here, the first and last characters in X_i are called the 0-th and (2^i-1)-th characters, respectively.
* S does not contain a string with length N+1 or more.



Here, a string A is a subsequence of another string B when there exists a sequence of integers t_1 < ... < t_{|A|} such that, for every i (1\leq i\leq |A|), the i-th character of A and the t_i-th character of B is equal.

Constraints

* 0 \leq N \leq 20
* X_i(0\leq i\leq N) is a string of length 2^i consisting of `0` and `1`.
* 1 \leq K \leq |S|
* K is an integer.

Input

Input is given from Standard Input in the following format:


N K
X_0
:
X_N


Output

Print the lexicographically smallest string among the longest strings that are subsequences of K or more different strings in S.

Examples

Input

3 4
1
01
1011
01001110


Output

10


Input

4 6
1
01
1011
10111010
1101110011111101


Output

100


Input

2 5
0
11
1111


Output

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
char str[1<<20];
int f[21][1<<20];
int nex[21][1<<20];
int main(){
//	freopen("F.in","r",stdin);
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;i++){
		scanf("%s",str);
		for(int j=0;j<(1<<i);j++)
			f[i][j]=str[j]-'0';
	}
	for(int i=1;i<=n;i++){
		for(int j=0;j<(1<<i);j++){
			int x=j>>(i-1)&1,p;
			for(p=i-1;p>=0&&(j>>p&1)==x;p--);
			nex[i][j]=p;
		}
	}
	int ans1=0,ans2=0;
	for(int i=0;i<=n;i++){
		for(int j=1;i+j<=n;j++){
			for(int k=0;k<(1<<(i+j));k++){
				f[i][k>>j]+=f[i+j][k];
				int d=nex[j][k&((1<<j)-1)];
				if(d!=-1){
					f[i+d+1][(k&((1<<(d+1))-1))|((k>>j)<<(d+1))]+=f[i+j][k];
				}
			}
		}
		for(int k=(1<<i)-1;k>=0;k--)
			if(f[i][k]>=m)
				ans1=i,ans2=k;
	}
	for(int i=ans1-1;i>=0;i--)
		printf("%d",ans2>>i&1);
	return 0;
}