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>
#define ll long long
using namespace std;
int f[22][1<<21],n,K;
char s[1<<21];
int main(){
	ios::sync_with_stdio(0);
	cin>>n>>K;
	for(int i=0,now=1;i<=n;i++,now<<=1){
		cin>>s;
		for(int j=0;j<1<<i;j++)
			f[i][j+now]=s[j]=='1';
	}
	for(int i=n;i;i--){
		for(int j=0;j<1<<n+1;j++){
			int k1=-1,k2=-1;
			for(int p=i-1;~p;p--)
				if(j>>p&1){k1=p;break;}
			for(int p=i-1;~p;p--)
				if(~j>>p&1){k2=p;break;}
			int tmp=f[i][j];
			f[0][j>>i]+=tmp;
			if(~k1){
				int now1=j&(1<<k1)-1,now2=(j>>i-1)|1;
				f[k1][now1|now2<<k1]+=tmp;
			}
			if(~k2){
				int now1=j&(1<<k2)-1,now2=(j>>i)<<1;
				f[k2][now1|now2<<k2]+=tmp;
			}
		}
	}
	int ans=1;
	for(int j=(1<<n+1)-1;j;j--){
		if(f[0][j]>=K){
			int now=n+1;
			while(~ans>>now&1&&~j>>now&1) now--;
			if(j>>now&1&&~ans>>now&1) ans=j;
			if(j>>now&1&&ans>>now&1&&ans>j) ans=j;
		}
	}
	int x=ans,now=n+1;
	while(~x>>now&1) now--;
	for(int i=now-1;~i;i--)
		cout<<(x>>i&1);
	cout<<'\n';
	return 0;
}