#include <bits/stdc++.h>
#define pb push_back
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a<b?a=b:a)
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
#define dec(i,s,t) for(int i=s;i>=(t);--i)
#define fore(i,v) for(int i=g[v],d=es[i].d;i;i=es[i].nxt,d=es[i].d)
using namespace std;

#define Pr(f,...) //fprintf(stderr,f,##__VA_ARGS__),fflush(stderr)

typedef long long ll;

const int N=21;

char str[1<<N];
int ans,len,n,req,f[N][1<<N],g[N][1<<N];

inline int mask(int x){ return (1<<x)-1; }

int main(int argc,char *argv[]){
#ifdef CURIOUSCAT
	freopen("dat.in","r",stdin);
	freopen("my.out","w",stdout);
#endif
	scanf("%d%d",&n,&req); 	
	rep(i,0,n){
		scanf("%s",str); 
		lop(s,0,(1<<i)){
			f[i][s]=str[s]-'0';
		}
	}
	g[0][0]=-1;
	rep(i,1,n){
		lop(s,0,(1<<i)){
			int b=s>>(i-1)&1,p=i;
			for(;p>=1&&(s>>(p-1)&1)==b;--p);
			g[i][s]=p;
		}	
	}
	rep(i,0,n){
		for(int j=1;i+j<=n;++j){
			int all=(1<<(i+j))-1;
			lop(s,0,all){
				f[i][s>>j]+=f[i+j][s];
				int p=g[j][s&mask(j)];
				if(p)f[i+p][((s>>j)<<p)+(s&mask(p))]+=f[i+j][s];
			}
		}
		dec(s,(1<<i)-1,0){
			if(f[i][s]>=req){
				len=i;
				ans=s;
			}
		}
	}
	dec(i,len-1,0)putchar((ans>>i&1)+'0'); 
	puts(""); 
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 pb push_back
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a<b?a=b:a)
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
#define dec(i,s,t) for(int i=s;i>=(t);--i)
#define fore(i,v) for(int i=g[v],d=es[i].d;i;i=es[i].nxt,d=es[i].d)
using namespace std;

#define Pr(f,...) //fprintf(stderr,f,##__VA_ARGS__),fflush(stderr)

typedef long long ll;

const int N=21;

char str[1<<N];
int ans,len,n,req,f[N][1<<N],g[N][1<<N];

inline int mask(int x){ return (1<<x)-1; }

int main(int argc,char *argv[]){
#ifdef CURIOUSCAT
	freopen("dat.in","r",stdin);
	freopen("my.out","w",stdout);
#endif
	scanf("%d%d",&n,&req); 	
	rep(i,0,n){
		scanf("%s",str); 
		lop(s,0,(1<<i)){
			f[i][s]=str[s]-'0';
		}
	}
	g[0][0]=-1;
	rep(i,1,n){
		lop(s,0,(1<<i)){
			int b=s>>(i-1)&1,p=i;
			for(;p>=1&&(s>>(p-1)&1)==b;--p);
			g[i][s]=p;
		}	
	}
	rep(i,0,n){
		for(int j=1;i+j<=n;++j){
			int all=(1<<(i+j))-1;
			lop(s,0,all){
				f[i][s>>j]+=f[i+j][s];
				int p=g[j][s&mask(j)];
				if(p)f[i+p][((s>>j)<<p)+(s&mask(p))]+=f[i+j][s];
			}
		}
		dec(s,(1<<i)-1,0){
			if(f[i][s]>=req){
				len=i;
				ans=s;
			}
		}
	}
	dec(i,len-1,0)putchar((ans>>i&1)+'0'); 
	puts(""); 
	return 0;
}

```