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;

#define CO const
#define IN inline
typedef long long int64;

template<class T> IN T read(){
	T x=0,w=1;char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-') w=-w;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*w;
}
template<class T> IN T read(T&x){
	return x=read<T>();
}
#define clz __builtin_clz

char str[(1<<20)+1];
int*F[1<<21],trans[1<<21][2];

int main(){
	int n=read<int>(),K=read<int>();
	F[1]=new int[1<<(n+1)];
	F[1][0]=0;
	for(int i=0;i<=n;++i){
		int p=1<<i;
		scanf("%s",str);
		for(int j=0;j<p;++j) F[1][p+j]=str[j]-'0'; // use a leading 1 to describe the length
		for(int j=0;j<p;++j) trans[p+j][1]=j;
		trans[2*p-1][0]=0; // doesn't exist
		for(int j=1;j<p;++j) trans[2*p-1-j][0]=j^((1<<(31-clz(j)))-1);
	}
	int ans=1;
	for(int i=1;i<=n;++i){
		int p=1<<i;
		for(int j=2*p-1;j>=p;--j){
			F[j]=new int[1<<(n-i+1)];
			memset(F[j],0,sizeof(int)<<(n-i+1));
			for(int k=1;k<1<<(n-i+2);++k) F[j][trans[k][j&1]]+=F[j>>1][k];
			F[j][0]=0;
			if(accumulate(F[j],F[j]+(1<<(n-i+1)),0)>=K) ans=j;
		}
	}
	for(int i=31-clz(ans)-1;i>=0;--i) printf("%d",ans>>i&1);
	puts("");
	return 0;
}