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;
const int N=21;
int n,k;
char s[(1<<N)+5];
vector<int> f[N][N];
int main(){
	scanf("%d%d",&n,&k);
	for(int i=0;i<=n;i++){
		scanf("%s",s);
		for(int j=0;j<=i;j++) f[i][j].resize(1<<i);
		for(int j=0;j<(1<<i);j++) if(s[j]=='1') f[i][0][j]=1;
	}
	for(int i=n;i>=1;i--){
		for(int j=0;j<i;j++){
			for(int s=0;s<(1<<i);s++){
				if(!f[i][j][s]) continue;
				int a=s&((1<<j)-1),b=s>>j;
				f[j][j][a]+=f[i][j][s];
				if(b) {
					int t=__builtin_ctz(b);
					f[i-t][j+1][(b>>t<<j)|a]+=f[i][j][s];
				}
				if(b!=(1<<(i-j))-1) {
					int t=__builtin_ctz(~b);
					f[i-t][j+1][(b>>t<<j)|a]+=f[i][j][s];
				}
			}
		}
		for(int s=0;s<(1<<i);s++){
			if(f[i][i][s]>=k){
				for(int j=i-1;j>=0;j--)
					if((s>>j)&1) putchar('1');
					else putchar('0');
				return 0;
			}
		}
	}
	return 0;
}