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>
#define meow(args...) fprintf(stderr, args)
int *f[1<<21], trans[1<<21][2];
char str[(1<<20)+1];
int main() {
	int n, k;
	assert(scanf("%d%d", &n, &k)==2);
	f[1]=new int[1<<(n+1)];
	f[1][0]=0;
	for(int i=0; i<=n; ++i) {
		int p=1<<i;
		assert(scanf("%s", str)==1);
		for(int j=0; j<p; ++j) f[1][p+j]=str[j]-'0';
		for(int j=0; j<p; ++j) trans[p+j][1]=j;
		trans[2*p-1][0]=0;
		for(int j=1; j<p; ++j) trans[2*p-j-1][0]=j^((1<<(31-__builtin_clz(j)))-1);
	}
	int ans=1;
	for(int i=1; i<=n; ++i) {
		int p=1<<i;
		for(int j=2*p; j-->p; ) {
			f[j]=new int[1<<(n-i+1)];
			memset(f[j], 0, sizeof(int)<<(n-i+1));
			for(int k=1; k<(1<<(n-i+2)); ++k) f[j][trans[k][j&1]]+=f[j>>1][k];
			f[j][0]=0;
			if(std::accumulate(f[j], f[j]+(1<<(n-i+1)), 0)>=k) ans=j;
		}
	}
	for(int i=31-__builtin_clz(ans); i--; ) printf("%d", ans>>i&1);
	puts("");
	return 0;
}