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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[22][(1<<20)+5];
int n,K;
int f[22][(1<<20)+5],g[22][(1<<20)+5];
inline int dp(int x,int y,int z){return f[x][y+(z<<(n-x))];}
int main(){
	scanf("%d%d",&n,&K);
	for(int i=0;i<=n;i++)scanf("%s",s[i]);
	for(int i=0;i<(1<<n);i++)
		if(s[n][i]=='1')g[0][i]=1;
	for(int i=n-1;i>=0;i--){
		swap(f,g);
		for(int k=0;k<(1<<i);k++){
			g[0][k]=f[0][k<<1]+f[0][(k<<1)|1]+(s[i][k]=='1');
			for(int j=1;i+j<=n;j++)
				for(int l=0;l<(1<<j);l++)
					if(l&(1<<(j-1)))
						g[j][ k+(l<<(n-j)) ]=dp(j,k<<1,l)+dp(j-1,(k<<1)|1,l^(1<<(j-1)));
					else
						g[j][ k+(l<<(n-j)) ]=dp(j-1,k<<1,l)+dp(j,(k<<1)|1,l);
		}
	}
	for(int i=n;i>0;i--){
		for(int j=0;j<(1<<i);j++)if(g[i][j<<(n-i)]>=K){
			for(int k=(i-1);k>=0;k--)putchar('0'+((j>>k)&1));
			puts("");
			return 0;
		}
	}
	return 0;
}