#include<bits/stdc++.h>
using namespace std;
#define cs const
#define pb push_back
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define bg begin
namespace IO{
	cs int RLEN=1<<20|1;
	char ibuf[RLEN],*ib,*ob;
	inline char gc(){
		(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
		return (ib==ob)?EOF:*ib++;
	}
	inline int read(){
		char ch=gc();
		int res=0;bool f=1;
		while(!isdigit(ch))f^=ch=='-',ch=gc();
		while(isdigit(ch))res=(res*10)+(ch^48),ch=gc();
		return f?res:-res;
	}
	inline void readstring(char *s){
		char ch=gc();int top=0;
		while(isspace(ch))ch=gc();
		while(!isspace(ch)&&ch!=EOF)s[top++]=ch,ch=gc();
	}
}using IO::read;
using IO::readstring;
template<typename tp>inline void chemx(tp &a,tp b){(a<b)?(a=b):0;}
template<typename tp>inline void chemn(tp &a,tp b){(a>b)?(a=b):0;}
cs int N=21,M=(1<<N)|5;
int lg[M],f[N][M],tr[M][2];
int n,K,lim;
char str[M];
void writesp(int x){
	if(x>1)writesp(x>>1);
	else return;
	putchar('0'+(x&1));
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),K=read(),lim=1<<n;
	for(int i=2;i<M;i++)lg[i]=lg[i>>1]+1;
	for(int i=0;i<=n;i++){
		readstring(str);
		for(int j=0;j<(1<<i);j++)f[i][(1<<i)^j]=(str[j]-'0');
	}
	for(int i=2;i<(lim<<1);i++){
		if((i>>(lg[i]-1))&1){
			tr[i][1]=i^(1<<lg[i]);
			tr[i][0]=tr[tr[i][1]][0];
		}else{
			tr[i][0]=i^(3<<(lg[i]-1));
			tr[i][1]=tr[tr[i][0]][1];
		}
	}
	for(int i=n;i;i--)
	for(int j=1<<i;j<(lim<<1);j++)if(f[i][j]){
		int p=j>>i,s=(j&((1<<i)-1))^(1<<i);
		if(tr[s][0])f[lg[tr[s][0]]][(((p<<1)+1)<<lg[tr[s][0]])^tr[s][0]]+=f[i][j];
		if(tr[s][1])f[lg[tr[s][1]]][(((p<<1)+0)<<lg[tr[s][1]])^tr[s][1]]+=f[i][j];
		f[0][p]+=f[i][j];
	}
	int mx=1;
	for(int i=2;i<(lim<<1);i++)if(f[0][i]>=K&&lg[i]>lg[mx])mx=i;
	writesp(mx);return 0;
}
### Prompt

In Cpp, your task is to solve 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>
using namespace std;
#define cs const
#define pb push_back
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define bg begin
namespace IO{
	cs int RLEN=1<<20|1;
	char ibuf[RLEN],*ib,*ob;
	inline char gc(){
		(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
		return (ib==ob)?EOF:*ib++;
	}
	inline int read(){
		char ch=gc();
		int res=0;bool f=1;
		while(!isdigit(ch))f^=ch=='-',ch=gc();
		while(isdigit(ch))res=(res*10)+(ch^48),ch=gc();
		return f?res:-res;
	}
	inline void readstring(char *s){
		char ch=gc();int top=0;
		while(isspace(ch))ch=gc();
		while(!isspace(ch)&&ch!=EOF)s[top++]=ch,ch=gc();
	}
}using IO::read;
using IO::readstring;
template<typename tp>inline void chemx(tp &a,tp b){(a<b)?(a=b):0;}
template<typename tp>inline void chemn(tp &a,tp b){(a>b)?(a=b):0;}
cs int N=21,M=(1<<N)|5;
int lg[M],f[N][M],tr[M][2];
int n,K,lim;
char str[M];
void writesp(int x){
	if(x>1)writesp(x>>1);
	else return;
	putchar('0'+(x&1));
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),K=read(),lim=1<<n;
	for(int i=2;i<M;i++)lg[i]=lg[i>>1]+1;
	for(int i=0;i<=n;i++){
		readstring(str);
		for(int j=0;j<(1<<i);j++)f[i][(1<<i)^j]=(str[j]-'0');
	}
	for(int i=2;i<(lim<<1);i++){
		if((i>>(lg[i]-1))&1){
			tr[i][1]=i^(1<<lg[i]);
			tr[i][0]=tr[tr[i][1]][0];
		}else{
			tr[i][0]=i^(3<<(lg[i]-1));
			tr[i][1]=tr[tr[i][0]][1];
		}
	}
	for(int i=n;i;i--)
	for(int j=1<<i;j<(lim<<1);j++)if(f[i][j]){
		int p=j>>i,s=(j&((1<<i)-1))^(1<<i);
		if(tr[s][0])f[lg[tr[s][0]]][(((p<<1)+1)<<lg[tr[s][0]])^tr[s][0]]+=f[i][j];
		if(tr[s][1])f[lg[tr[s][1]]][(((p<<1)+0)<<lg[tr[s][1]])^tr[s][1]]+=f[i][j];
		f[0][p]+=f[i][j];
	}
	int mx=1;
	for(int i=2;i<(lim<<1);i++)if(f[0][i]>=K&&lg[i]>lg[mx])mx=i;
	writesp(mx);return 0;
}
```