#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
int id[21][21];
int rev[1<<20];
int S,T;
int n,t,k;
int getid(int i,int x,int j,int y){
	return id[i][j]+y+(x<<j);
}
const int N=1e8+10;
int dp[N];
void move(int x,int y){
	if (!x||!y)return;
	dp[y]+=dp[x];
}
int main(){
	cin>>n>>k;
	for (int i=0;i<=n;i++){
		for (int j=0;j<=n-i;j++){
			id[i][j]=t+1;
			t+=1<<i+j;
		}
	}
	for (int i=0;i<=n;i++){
		scanf("\n");
		for (int s=0;s<1<<i;s++)dp[getid(0,0,i,s)]=getchar()-'0';
	}
	for (int j=n;j;j--)
	for (int i=0;i+j<=n;i++)
	for (int x=0;x<1<<i;x++)
	for (int y=0;y<1<<j;y++){
		int p=getid(i,x,j,y),z=y&1,x0=x,y0=y;
		move(p,getid(i,x,0,0));
		move(p,getid(i+1,x<<1|z,j-1,y>>1));
		for (int d=0;d<j;d++)
			if ((1<<d&y)>>d!=z){
				move(p,getid(i+1,x<<1|(z^1),j-d-1,y>>d+1));
				break;
			}
	}
	int ans=0,a1=0;
	for (int i=1;i<=n;i++){
		for (int s=0;s<1<<i;s++){
			rev[s]=(rev[s>>1]>>1)+(s&1)*(1<<i-1);
			if (dp[getid(i,s,0,0)]>=k){
				if (a1<i)ans=rev[s];
				ans=min(ans,rev[s]);
				a1=i;
			}
		}
	}
	for (int i=a1-1;~i;i--)putchar('0'+(bool)(ans&(1<<i)));
}
### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
int id[21][21];
int rev[1<<20];
int S,T;
int n,t,k;
int getid(int i,int x,int j,int y){
	return id[i][j]+y+(x<<j);
}
const int N=1e8+10;
int dp[N];
void move(int x,int y){
	if (!x||!y)return;
	dp[y]+=dp[x];
}
int main(){
	cin>>n>>k;
	for (int i=0;i<=n;i++){
		for (int j=0;j<=n-i;j++){
			id[i][j]=t+1;
			t+=1<<i+j;
		}
	}
	for (int i=0;i<=n;i++){
		scanf("\n");
		for (int s=0;s<1<<i;s++)dp[getid(0,0,i,s)]=getchar()-'0';
	}
	for (int j=n;j;j--)
	for (int i=0;i+j<=n;i++)
	for (int x=0;x<1<<i;x++)
	for (int y=0;y<1<<j;y++){
		int p=getid(i,x,j,y),z=y&1,x0=x,y0=y;
		move(p,getid(i,x,0,0));
		move(p,getid(i+1,x<<1|z,j-1,y>>1));
		for (int d=0;d<j;d++)
			if ((1<<d&y)>>d!=z){
				move(p,getid(i+1,x<<1|(z^1),j-d-1,y>>d+1));
				break;
			}
	}
	int ans=0,a1=0;
	for (int i=1;i<=n;i++){
		for (int s=0;s<1<<i;s++){
			rev[s]=(rev[s>>1]>>1)+(s&1)*(1<<i-1);
			if (dp[getid(i,s,0,0)]>=k){
				if (a1<i)ans=rev[s];
				ans=min(ans,rev[s]);
				a1=i;
			}
		}
	}
	for (int i=a1-1;~i;i--)putchar('0'+(bool)(ans&(1<<i)));
}
```