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<cstring>
#include<cstdio>
using namespace std;
#define N 2097153
int n,K,f[N],g[N],s[21][N>>1],len,ans=-1,ans1=0,le[24],all;char sc[N];
int main(){
	for(int i=0;i<=20;i++)le[i]=(1<<i)-1;all=(1<<21)-1;
	scanf("%d%d",&n,&K);int i,j,k,v,v1,v2;
	for(i=0;i<=n;i++){scanf("%s",sc);for(k=0;k<(1<<i);k++)if(sc[k]=='1'){v=1;for(j=0;j<i;j++)v=(v<<1)|((k>>j)&1);++f[v];}}
	for(i=0;i<=n;i++)
	{
		memset(g,0,sizeof(g));for(j=(1<<i);j<N;j++)s[i][j&le[i]]+=f[j];
		for(j=1<<(i+1);j<N;j++)
		if(v=f[j])
		{
			v1=j>>i;v1=v1>>(__builtin_ctz(v1)+1);if(v1)g[(v1<<(i+1))|(((j&le[i])<<1)|1)]+=v;
			v2=j>>i;v2=v2>>(__builtin_ctz(v2^all)+1);if(v2)g[(v2<<(i+1))|((j&le[i])<<1)]+=v;
		}
		memcpy(f,g,sizeof(g));
	}
	for(i=0;i<=n;i++)
	for(j=0;j<(1<<n);j++)if(s[i][j]>=K){if(i==ans)ans1=min(ans1,j);if(i>ans){ans=i;ans1=j;}}
	for(i=ans-1;i>=0;i--)printf("%d",(ans1>>i)&1);return 0;
}