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<algorithm>
#include<cstdio>
using namespace std;
int n,K,sl,fh,f[1<<21][22];
char X[1<<20];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
void print(int s,int i)
{
	for(--i;~i;putchar(((s>>i)&1)+'0'),--i);
	puts("");
}
int main()
{
	n=rd();K=rd();int x,s1,s2,res;
	for(int i=0;i<=n;++i)
	{
		scanf("%s",X);
		for(int j=0;j<1<<i;++j)
			f[(1<<i)|j][i]=X[j]=='1';
	}
	for(int p,i=n;i;--i)
		for(int j=0;j<1<<i;++j)
			for(int k=i;k;--k)
				if((res=f[x=(1<<i)|j][k]))
				{
					s1=x>>k;s2=x^(s1<<k);
					if(k) f[s1][0]+=res;
					if(s2)
					{
						p=k-1;for(;(~p)&&!((s2>>p)&1);--p);
						f[(s1<<1|1)<<p|(s2&((1<<p)-1))][p]+=res;
					}
					if((~s2)&((1<<k)-1))
					{
						p=k-1;for(;(~p)&&((s2>>p)&1);--p);
						f[(s1<<1)<<p|(s2&((1<<p)-1))][p]+=res;
					}
				}
	for(int i=n;i;--i)
		for(int j=0;j<1<<i;++j)
			if(f[(1<<i)|j][0]>=K)
				return print(j,i),0;
	return 0;
}