#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
cs int N = 1 << 20 | 5, M = 22;
int n, K, Ans, Alen=-1; char str[N];
int *dp[M][M];
struct dat{
	int S, len; dat(int _S=0, int l=0){ S=_S; len=l; }
	int fnd(int z){ 
		for(int i=len-1;~i;i--)
		if((S>>i&1)==z) return i+1; return -1;
	} void trans(int k){ --k; len=k; S=S&((1<<k)-1); }
};
void Add(dat a, dat b, int coe){
	int x = a.len + b.len, y = a.len;
	int z = (a.S << b.len) + b.S; 
	dp[x][y][z] += coe;
}
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	scanf("%d%d",&n,&K);
	for(int i=0; i<=n; i++)
	for(int j=0; j<=i; j++){
		dp[i][j] = new int[1<<i];
		memset(dp[i][j],0,sizeof(int)*(1<<i));
	}
	for(int i=0; i<=n; i++){
		scanf("%s",str); int m=1<<i;
		for(int j=0; j<m; j++) 
		if(str[j]=='1') Add(dat(0,0),dat(j,i),1);
	} 
	for(int i=0; i<=n; i++)
	for(int j=i+1,coe; j<=n; j++)
	for(int k=0,m=1<<j; k<m; k++)
	if(coe=dp[j][i][k]){
		dat A(k>>(j-i),i), na = A;
		dat B(k&((1<<(j-i))-1),j-i), nb = B;
		int a = B.fnd(0), b = B.fnd(1);
		if(~a){
			na.S = na.S << 1, ++na.len;
			nb.trans(a); Add(na,nb,coe); 
		} na = A; nb = B;
		if(~b){
			na.S = na.S << 1 | 1; ++na.len;
			nb.trans(b); Add(na,nb,coe);
		} 
	} 
	static int ct[M][N];
	for(int i=0; i<=n; i++)
	for(int j=i; j<=n; j++)
	for(int k=0; k<(1<<j); k++){
		int S=k>>(j-i); ct[i][S]+=dp[j][i][k];
	}
	for(int i=n; i>=0; i--)
	for(int j=0; j<(1<<i); j++) 
	if(ct[i][j]>=K){ for(int t=i-1;~t;t--) cout<<(j>>t&1); 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 cs const
#define pb push_back
using namespace std;
cs int N = 1 << 20 | 5, M = 22;
int n, K, Ans, Alen=-1; char str[N];
int *dp[M][M];
struct dat{
	int S, len; dat(int _S=0, int l=0){ S=_S; len=l; }
	int fnd(int z){ 
		for(int i=len-1;~i;i--)
		if((S>>i&1)==z) return i+1; return -1;
	} void trans(int k){ --k; len=k; S=S&((1<<k)-1); }
};
void Add(dat a, dat b, int coe){
	int x = a.len + b.len, y = a.len;
	int z = (a.S << b.len) + b.S; 
	dp[x][y][z] += coe;
}
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	scanf("%d%d",&n,&K);
	for(int i=0; i<=n; i++)
	for(int j=0; j<=i; j++){
		dp[i][j] = new int[1<<i];
		memset(dp[i][j],0,sizeof(int)*(1<<i));
	}
	for(int i=0; i<=n; i++){
		scanf("%s",str); int m=1<<i;
		for(int j=0; j<m; j++) 
		if(str[j]=='1') Add(dat(0,0),dat(j,i),1);
	} 
	for(int i=0; i<=n; i++)
	for(int j=i+1,coe; j<=n; j++)
	for(int k=0,m=1<<j; k<m; k++)
	if(coe=dp[j][i][k]){
		dat A(k>>(j-i),i), na = A;
		dat B(k&((1<<(j-i))-1),j-i), nb = B;
		int a = B.fnd(0), b = B.fnd(1);
		if(~a){
			na.S = na.S << 1, ++na.len;
			nb.trans(a); Add(na,nb,coe); 
		} na = A; nb = B;
		if(~b){
			na.S = na.S << 1 | 1; ++na.len;
			nb.trans(b); Add(na,nb,coe);
		} 
	} 
	static int ct[M][N];
	for(int i=0; i<=n; i++)
	for(int j=i; j<=n; j++)
	for(int k=0; k<(1<<j); k++){
		int S=k>>(j-i); ct[i][S]+=dp[j][i][k];
	}
	for(int i=n; i>=0; i--)
	for(int j=0; j<(1<<i); j++) 
	if(ct[i][j]>=K){ for(int t=i-1;~t;t--) cout<<(j>>t&1); return 0; }
}
```