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 MAXN 2097152
using namespace std;
int f[MAXN+5][21];
char s[MAXN+5];
int main (){
	int i,j,k,l,n,tar;
	scanf ("%d",&n);
	scanf ("%d",&tar);
	for (i=0;i<=n;i++)
	{scanf ("%s",s);
	for (j=0;j<(1<<i);j++)
	{if (s[j]=='1') f[(1<<i)+j][0]=1;
	}
	}
	for (i=0;i<=20;i++)
	{for (j=20-i;j>=0;j--)
	{for (l=0;l<(1<<(i+j));l++)
	{int id=(1<<(i+j))+l,o=i;
	if (!f[id][o]) continue;
	int s1=(l>>j),s2=l&((1<<j)-1);
	int l1=i,l2=j,f0=-1,f1=-1;
	if (j) f[(1<<i)+s1][i]+=f[id][o];
	for (k=j-1;k>=0;k--)
	{if (!(s2&(1<<k))) {f0=k;break;}
	}
	for (k=j-1;k>=0;k--)
	{if (s2&(1<<k)) {f1=k;break;}
	}
	if (f0!=-1)
	{int rm=s2&((1<<f0)-1);
	int nta=(1<<(i+1+f0))+(s1<<(f0+1))+rm;
	f[nta][i+1]+=f[id][o];
	}
	if (f1!=-1)
	{int rm=s2&((1<<f1)-1);
	int nta=(1<<(i+1+f1))+(((s1<<1)+1)<<f1)+rm;
	f[nta][i+1]+=f[id][o];
	}
	}
	}
	}
	for (i=20;i>=0;i--)
	{for (j=0;j<(1<<i);j++)
	{if (f[(1<<i)+j][i]>=tar)
	{for (l=i-1;l>=0;l--)
	{if (j&(1<<l)) {putchar('1');}
	else {putchar('0');}
	}
	return 0;
	}
	}
	}
	return 0;
}