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;
typedef long long ll;
const int N=25;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,m,ans,ansl;
int F[1<<21],G[1<<21];
int *f[N],*g[N];
char ch[1<<20];
int main(){
	n=read();m=read();
	for (int i=0;i<=n;i++){
		f[i]=F+(1<<i);g[i]=G+(1<<i);
		scanf(" %s",ch);
		for (int j=0;j<(1<<i);j++){
			f[i][j]=ch[j]-'0';
			int fl=(j>>(i-1))&1,k=i-1;
			for (;(k>=0)&&(((j>>k)&1)==fl);k--);
			g[i][j]=k+1;
		}
	}
	for (int i=0;i<=n;i++)
	for (int j=1;i+j<=n;j++)
	for (int k=0;k<(1<<(i+j));k++){
		int kl=k>>j,kr=k&((1<<j)-1),t=g[j][kr];
		f[i][k>>j]+=f[i+j][k];
		if (t) f[i+t][(kl<<t)|(kr&((1<<t)-1))]+=f[i+j][k];
	}
	ans=-1;
	for (int i=0;i<=n;i++)
	for (int k=(1<<i)-1;k>=0;k--)
	if (f[i][k]>=m){ans=i;ansl=k;}
	for (int i=ans-1;i>=0;i--) putchar('0'+((ansl>>i)&1));
	puts("");
	return 0;
}