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,m,f[N][1<<N];
char s[1<<N];
int main(){
	scanf("%d%d",&n,&m);
	for (int i=0;i<=n;i++){
		scanf("%s",s);
		for (int j=0;j<1<<i;j++)f[0][j<<n-i+1|1<<n-i]=s[j]-'0';
	}
	for (int i=0;i<n;i++)
		for (int j=0;j<1<<n+1;j++)
			if (f[i][j]){
				int a=j&(1<<i)-1,b=j^a,c=__builtin_ctz(b);
				b^=1<<c;
				if (b)f[i+1][b|1<<i|a]+=f[i][j];
				if (b+(1<<c+1)<(1<<n+1)){
					int nc=__builtin_ctz(~b>>c+1<<c+1);
					f[i+1][b>>nc<<nc|1<<nc|a]+=f[i][j];
				}
			}
	for (int i=n;i>=0;i--)
		for (int j=0;j<1<<i;j++){
			int res=0;
			for (int k=0;k<1<<n+1-i;k++)res+=f[i][k<<i|j];
			if (res>=m){
				for (int k=i-1;k>=0;k--)putchar('0'+(j>>k&1));
				return 0;
			}
		}
}