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;
vector<int> f[N][N];
int n,K; char s[1<<N];
int main()
{
	scanf("%d%d",&n,&K);
	for(int i=0;i<=n;++i)
	{
		scanf("%s",s);
		for(int j=0;i+j<=n;++j) f[i][j].resize(1<<i+j);
		for(int j=0;s[j];++j) f[i][0][j]=(s[j]-'0');
	}
	for(int i=n;i;--i)
		for(int j=0;i+j<=n;++j)
			for(int k=0;k<(1<<i+j);++k) if(f[i][j][k])
			{
				int a=k>>j, b=(j?k&((1<<j)-1):0);
				f[0][j][b]+=f[i][j][k];
				if(a)
				{
					int p=32-__builtin_clz(a),nb=j?(b<<1|1):1;
					f[i-(i-p+1)][j+1][(a&((1<<p-1)-1))<<(j+1)|nb]+=f[i][j][k];
				}
				if(a!=(1<<i)-1)
				{
					unsigned tmp=(unsigned)a<<(32-i);
					int p=i-__builtin_clz(~tmp),nb=j?(b<<1):0;
					f[i-(i-p+1)][j+1][(a&((1<<p-1)-1))<<(j+1)|nb]+=f[i][j][k];
				}
			}
	for(int i=n;i;--i)
		for(int k=0;k<(1<<i);++k) if(f[0][i][k]>=K)
		{
			for(int j=i-1;~j;--j) putchar((k>>j&1)+'0');
			return 0;
		}
}