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
#define re register
#define ull unsigned ll
using namespace std;
inline int read(){
	int s=0,t=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')t=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*t;
}
const int N=21;
char s[1<<N];
int n,m,U,Num,Ans,p[1<<N],tr[1<<N][2],f[N][1<<N];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	n=read(),m=read(),U=(1<<n+1)-1;
	for(int i=0;i<=n;i++){
		scanf("%s",s);
		for(int x=0;x<1<<i;x++)f[i][1<<i^x]=s[x]-'0';
	}
	for(int S=2;S<=U;S++)p[S]=p[S>>1]+1;
	for(int S=2;S<=U;S++){
		if(S>>(p[S]-1)&1){
			tr[S][1]=S^(1<<p[S]);
			tr[S][0]=tr[tr[S][1]][0];
		}
		else{
			tr[S][0]=S^(3<<p[S]-1);
			tr[S][1]=tr[tr[S][0]][1];
		}
	}// A|B
	for(int i=n;i>=1;i--){
		for(int S=1<<i;S<=U;S++){
			if(f[i][S]==0)continue;
			int A=S>>i,B=(S&((1<<i)-1))^(1<<i);
			for(int t=0;t<2;t++)if(tr[B][t]){
				int k=p[tr[B][t]];
				f[k][(A<<1^t^1)<<k^tr[B][t]]+=f[i][S];
			}f[0][A]+=f[i][S];
		}
	}Ans=1,Num=n;
	for(int S=2;S<=U;S++)if(f[0][S]>=m&&p[S]>p[Ans])Ans=S;
	while(!(Ans>>Num&1))Num--;
	while(Num--)printf("%d",Ans>>Num&1);
	return 0;
}