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<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<map>
const int N=1048580,M=21;
using namespace std;
int n,m,len,ans;
int f[M][N],g[M][N];
char s[N];

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;i++){
		scanf("%s",s);
		for(int j=0;j<(1<<i);j++)f[i][j]=s[j]-'0';//cout<<f[i][j]<<" ";;puts("");
	}
	for(int i=1;i<=n;i++)for(int j=0;j<(1<<i);j++){
		int b=(j>>(i-1))&1,r=0;
		for(;r<i&&((j>>(i-r-1))&1)==b;r++);
		g[i][j]=r==i?-1:r;
	}
	for(int i=0;i<=n;i++){
		for(int j=1;j<=n-i;j++)
		for(int k=0;k<(1<<(i+j));k++){
			f[i][k>>j]+=f[i+j][k];
			int t=g[j][k&((1<<j)-1)];
			if(t==-1)continue;
			f[i+j-t][((k>>j)<<(j-t))|(k&((1<<(j-t))-1))]+=f[i+j][k];
		}
	//	cout<<i<<endl;
		for(int j=(1<<i)-1;j>=0;j--){
		//	cout<<f[i][j]<<endl;
		//	cout<<i<<" "<<j<<endl;
			if(f[i][j]>=m){
				len=i;ans=j;
			}
		}
	}
	for(int i=len-1;i>=0;i--){
		printf("%d",(ans>>i)&1);
	}
	return 0;
}