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>
const int S=11e5;
int n,m,nw,f[2][23][S],i,j,k,ml,ans,x,y,z,nxt[S][23];
char c[S];
int main(){
	scanf("%d%d%s",&n,&m,c);
	for(i=1;i<=n;++i){
		scanf("%s",c);
		for(j=0;j<1<<i;++j)f[0][i][j]=c[j]=='1';
	}
	for(i=0;i<S;++i)for(j=19,nxt[i][20]=20;j>=0;--j)
		nxt[i][j]=(i>>j&1)==(i>>j+1&1)?nxt[i][j+1]:j+1;
	for(i=1;i<=n;++i){
		nw^=1;
		for(j=n;j>=i;--j)memcpy(f[nw][j],f[!nw][j],4<<j);
		for(j=n;j>i;--j)for(k=0;k<1<<j;++k){
			if(x=f[!nw][j][k]){
				if(y=nxt[k][i-1],y<j)f[nw][j-(y-(i-1))][(k&((1<<i-1)-1))|(k>>y<<i-1)]+=x;
			}
			f[nw][i][k&((1<<i)-1)]+=f[nw][j][k];
		}
		for(j=(1<<i)-1;j>=0;--j)if(f[nw][i][j]>=m)ans=j,ml=i;
	}
	for(k=ml-1;k>=0;--k)putchar('0'+(ans>>k&1));puts("");
	return 0;
}