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;

int *f[21][21];
int sum[21][1100010],nxt[1100010][21][2],n,k;
char s[21][1100010];

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

int main()
{
	n=rd();k=rd();
	for (int i=0;i<=n;i++) scanf("%s",s[i]+1);
	for (int i=0;i<=n;i++) for (int j=0;j<=i;j++)
	{
		f[i][j]=new int[1<<i];
		memset(f[i][j],0,sizeof(int)*(1<<i));
	}
	for (int i=n;i;i--)
	{
		for (int j=0;j<(1<<i);j++) if (s[i][j+1]=='1') f[i][0][j]++;
		for (int j=0;j<(1<<i);j++)
		{
			nxt[j][i][0]=nxt[j][i][1]=i;
			for (int k=i-1;~k;k--)
			{
				nxt[j][k][0]=nxt[j][k+1][0];
				nxt[j][k][1]=nxt[j][k+1][1];
				if ((j>>k)&1) nxt[j][k][1]=k;
				else nxt[j][k][0]=k;
			}
		}
		for (int k=0;k<=i;k++) for (int j=0;j<(1<<i);j++)
		{
			if (k) sum[k][j&((1<<k)-1)]+=f[i][k][j];
			int hh=nxt[j][k][0];
			if (hh<i) f[i-(hh-k)][k+1][(j&((1<<k)-1))+(j>>(hh+1)<<(k+1))]+=f[i][k][j];
			hh=nxt[j][k][1];
			if (hh<i) f[i-(hh-k)][k+1][(j&((1<<k)-1))+(j>>(hh+1)<<(k+1))+(1<<k)]+=f[i][k][j];
		}
		for (int j=0;j<(1<<i);j++) if (sum[i][j]>=k)
		{
			for (int l=i-1;~l;l--) putchar(((j>>l)&1)+'0');
			putchar('\n');
			return 0;
		}
	}
	return 0;
}