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 For(i,j,k) for (int i=j;i<=k;i++)
using namespace std;
const int N=1048580,M=21;
int n,m,len,ans;
int f[M][N],g[M][N];
char s[N];
int main(){
	scanf("%d%d",&n,&m);
	For(i,0,n){
		scanf("%s",s);
		For(j,0,(1<<i)-1)
			f[i][j]=s[j]-'0';
	}
	For(i,1,n) For(j,0,(1<<i)-1){
		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(i,0,n){
		For(j,1,n-i) For(k,0,(1<<(i+j))-1){
			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];
		}
		for (int j=(1<<i)-1;j>=0;j--)
			if (f[i][j]>=m){
				len=i; ans=j;
			}
	}
	for (int i=len-1;i>=0;i--)
		printf("%d",(ans>>i)&1);
}