#include<bits/stdc++.h>
#define title "title"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<int,int>
#define vint vector<int>
#define pb push_back
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
int read(){
	int g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const int N=20;
int n,m,hav[N+5][2<<N],s[2<<N],ne[2<<N][2],ans;
signed main(){
	n=read(),m=read();
	for(int i=0;i<=n;i++)for(int x=0;x<1<<i;x++)scanf("%1d",&hav[i][1<<i^x]);
	s[0]=-1;for(int i=1;i<2<<n;i++)s[i]=s[i>>1]+1;
	for(int i=2;i<2<<n;i++){
		if(i>>(s[i]-1)&1)ne[i][1]=i^1<<s[i],ne[i][0]=ne[ne[i][1]][0];
		else ne[i][0]=i^3<<(s[i]-1),ne[i][1]=ne[ne[i][0]][1];
	}
	for(int i=n;i>=1;i--)for(int sta=1<<i;sta<2<<n;sta++){
		if(!hav[i][sta])continue;
		int re=sta>>i,to=(sta&((1<<i)-1))^1<<i;
		if(ne[to][0])hav[s[ne[to][0]]][(re<<1^1)<<s[ne[to][0]]^ne[to][0]]+=hav[i][sta];
		if(ne[to][1])hav[s[ne[to][1]]][(re<<1^0)<<s[ne[to][1]]^ne[to][1]]+=hav[i][sta];
		hav[0][re]+=hav[i][sta];
	}ans=1;
	for(int sta=2;sta<2<<n;sta++)if(hav[0][sta]>=m&&s[sta]>s[ans])ans=sta;
	int j=n;while(~ans>>j&1)j--;while(j--)cout<<(ans>>j&1);return signed();
}
### 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<bits/stdc++.h>
#define title "title"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<int,int>
#define vint vector<int>
#define pb push_back
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
int read(){
	int g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const int N=20;
int n,m,hav[N+5][2<<N],s[2<<N],ne[2<<N][2],ans;
signed main(){
	n=read(),m=read();
	for(int i=0;i<=n;i++)for(int x=0;x<1<<i;x++)scanf("%1d",&hav[i][1<<i^x]);
	s[0]=-1;for(int i=1;i<2<<n;i++)s[i]=s[i>>1]+1;
	for(int i=2;i<2<<n;i++){
		if(i>>(s[i]-1)&1)ne[i][1]=i^1<<s[i],ne[i][0]=ne[ne[i][1]][0];
		else ne[i][0]=i^3<<(s[i]-1),ne[i][1]=ne[ne[i][0]][1];
	}
	for(int i=n;i>=1;i--)for(int sta=1<<i;sta<2<<n;sta++){
		if(!hav[i][sta])continue;
		int re=sta>>i,to=(sta&((1<<i)-1))^1<<i;
		if(ne[to][0])hav[s[ne[to][0]]][(re<<1^1)<<s[ne[to][0]]^ne[to][0]]+=hav[i][sta];
		if(ne[to][1])hav[s[ne[to][1]]][(re<<1^0)<<s[ne[to][1]]^ne[to][1]]+=hav[i][sta];
		hav[0][re]+=hav[i][sta];
	}ans=1;
	for(int sta=2;sta<2<<n;sta++)if(hav[0][sta]>=m&&s[sta]>s[ans])ans=sta;
	int j=n;while(~ans>>j&1)j--;while(j--)cout<<(ans>>j&1);return signed();
}
```