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 ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=(1<<20)+5;
int n,K,ans1,ans2;
int f[21][N];
int nx[21][N];
char s[N];
int main(){
	scanf("%d%d",&n,&K);
	For(i,0,n){
		scanf("%s",s);
		For(j,0,(1<<i)-1)
			f[i][j]=s[j]-'0';
	}
	For(i,0,n)
		For(j,0,(1<<i)-1){
			int fl=(j&1),p=0;
			for (;p<i&&!(((j>>p)&1)^fl);++p);
			nx[i][j]=(p>=i?-1:p);
		}
	For(i,0,n)
		For(j,1,n-i)
			For(k,0,(1<<(i+j))-1){
				f[i][k&((1<<i)-1)]+=f[i+j][k];
				int t=nx[j][k>>i];
				if (t==-1) continue;
				f[i+j-t][(k&((1<<i)-1))|((k>>(i+t))<<i)]+=f[i+j][k];
			}
	ans1=1<<30;
	For(i,0,n)
		For(j,0,(1<<i)-1){
			if (f[i][j]>=K)
				if (i>ans2||(i==ans2&&j<=ans1))
					ans1=j,ans2=i;
		}
	Rep(i,ans2-1,0)
		putchar('0'+((ans1>>i)&1));
}