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 gc getchar()
using namespace std;
int n,m,f[21][1<<20],g[21][1<<20];
char str[1<<21];
int read()
{
	int x=1;
	char ch;
	while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;
	int s=ch-'0';
	while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';
	return s*x;
}
int main()
{
	n=read(),m=read();
	for (int i=0;i<=n;i++)
	{
		scanf("%s",&str);
		for (int j=0;j<(1<<i);j++)
			f[i][j]=(str[j]=='1');
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=0;j<(1<<i);j++)
		{
			int fir=j>>(i-1)&1,pos=0;
			while (pos<i&&(j>>(i-pos-1)&1)==fir) pos++;
			g[i][j]=pos==i?-1:pos;
		}
	}
	int ans=0,len=0;
	for (int i=0;i<=n;i++)
	{
		for (int j=1;i+j<=n;j++)
		{
			for (int k=0;k<(1<<(i+j));k++)
			{
				f[i][k>>j]+=f[i+j][k];
				int t=g[j][k&((1<<j)-1)];
				if (t!=-1) f[i+j-t][k>>j<<(j-t)|(k&((1<<(j-t))-1))]+=f[i+j][k];
			}
		}
		for (int j=(1<<i)-1;~j;j--)
			if (f[i][j]>=m) ans=j,len=i;
	}
	for (int i=len;i;i--)
		putchar((ans>>(i-1)&1)+'0');
	puts("");
	return 0;
}