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 maxn=(1<<21)+10;
int n,m,f[21][maxn],lg[maxn],tr[maxn][2];
int main()
{
	scanf("%d%d",&n,&m);
	for(int j=0;j<=n;j++)
		for(int i=0;i<(1<<j);i++)
			scanf("%1d",&f[j][i^(1<<j)]);
	lg[0]=-1;
	for(int i=1;i<(1<<(n+1));i++) lg[i]=lg[i>>1]+1;
	for(int i=2;i<(1<<(n+1));i++)
		if((i>>(lg[i]-1))&1)
			tr[i][1]=i^(1<<lg[i]),tr[i][0]=tr[tr[i][1]][0];
		else
			tr[i][0]=i^(3<<(lg[i]-1)),tr[i][1]=tr[tr[i][0]][1];
	for(int j=n;j;j--)
		for(int i=(1<<j);i<(1<<(n+1));i++)
		{
			int k = i >> j, l = (i & ((1 << j) - 1)) ^ (1 << j);
			if(tr[l][0]) f[lg[tr[l][0]]][(((k << 1) ^ 1) << lg[tr[l][0]]) ^ tr[l][0]] += f[j][i];
			if(tr[l][1]) f[lg[tr[l][1]]][(((k << 1) ^ 0) << lg[tr[l][1]]) ^ tr[l][1]] += f[j][i];
			f[0][k] += f[j][i];
		}
	int ans=1;
	for(int i=2;i<(1<<(n+1));i++)
		if(f[0][i]>=m && lg[i]>lg[ans]) ans=i;
	int j=n;
	while((~ans>>j)&1) j--;
	while(j--) putchar(((ans>>j)&1)+'0');
	putchar('\n');
	return 0;
}