#include<bits/stdc++.h>

#define For(i,x,y) for (int i=x;i<y;i++)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define lf else if

#define dprintf(...) fprintf(stderr,__VA_ARGS__)
using namespace std;

typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> Vi;

int IN(){
	int c,f,x;
	while (!isdigit(c=getchar())&&c!='-');c=='-'?(f=1,x=0):(f=0,x=c-'0');
	while (isdigit(c=getchar())) x=(x<<1)+(x<<3)+c-'0';return !f?x:-x;
}

const int N=21;

int n,k,t;
int f[2][N][1<<N];
int Lg[1<<N],rev[1<<N],sum[1<<N];
pii trs[N][1<<N][2],ans;
char s[1<<N];

int main(){
	//freopen("01.txt","r",stdin);
	For(i,1,1<<N) Lg[i]=Lg[i>>1]+1;
	n=IN(),k=IN();
	For(i,0,n+1) For(x,0,1<<i){
		if (x==(1<<i)-1){
			trs[i][x][0]=mp(-1,-1);
		} else{
			int w=Lg[(x+1)&-(x+1)];
			trs[i][x][0]=mp(i-w,x>>w);
		}
		if (x==0){
			trs[i][x][1]=mp(-1,-1);
		} else{
			int w=Lg[x&-x];
			trs[i][x][1]=mp(i-w,x>>w);
		}
	}
	For(i,0,n+1){
		scanf("%s",s);
		For(j,1,1<<i) rev[j]=(rev[j>>1]>>1)|(j&1)<<i-1;
		For(j,0,1<<i) if (s[j]=='1') f[t^1][i][rev[j]]=1;
	}
	For(i,0,n+1){
		memset(sum,0,sizeof(sum));
		t^=1;
		For(j,0,n-i+1) For(k,0,1<<i+j)
			if (f[t][j][k]){
				int x=k>>i,y=k&(1<<i)-1;
				For(w,0,2){
					pii P=trs[j][x][w];
					if (P.fi==-1) continue;
					f[t^1][P.fi][P.se<<(i+1)|y<<1|w]+=f[t][j][k];
				}
				sum[y]+=f[t][j][k];
				f[t][j][k]=0;
			}
		For(x,0,1<<i) if (sum[x]>=k){
			ans=mp(i,x);
			break;
		}
	}
	for (int i=ans.fi-1;~i;i--) printf("%d",ans.se>>i&1);
	puts("");
}
### Prompt

Please create a solution in CPP 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 For(i,x,y) for (int i=x;i<y;i++)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define lf else if

#define dprintf(...) fprintf(stderr,__VA_ARGS__)
using namespace std;

typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> Vi;

int IN(){
	int c,f,x;
	while (!isdigit(c=getchar())&&c!='-');c=='-'?(f=1,x=0):(f=0,x=c-'0');
	while (isdigit(c=getchar())) x=(x<<1)+(x<<3)+c-'0';return !f?x:-x;
}

const int N=21;

int n,k,t;
int f[2][N][1<<N];
int Lg[1<<N],rev[1<<N],sum[1<<N];
pii trs[N][1<<N][2],ans;
char s[1<<N];

int main(){
	//freopen("01.txt","r",stdin);
	For(i,1,1<<N) Lg[i]=Lg[i>>1]+1;
	n=IN(),k=IN();
	For(i,0,n+1) For(x,0,1<<i){
		if (x==(1<<i)-1){
			trs[i][x][0]=mp(-1,-1);
		} else{
			int w=Lg[(x+1)&-(x+1)];
			trs[i][x][0]=mp(i-w,x>>w);
		}
		if (x==0){
			trs[i][x][1]=mp(-1,-1);
		} else{
			int w=Lg[x&-x];
			trs[i][x][1]=mp(i-w,x>>w);
		}
	}
	For(i,0,n+1){
		scanf("%s",s);
		For(j,1,1<<i) rev[j]=(rev[j>>1]>>1)|(j&1)<<i-1;
		For(j,0,1<<i) if (s[j]=='1') f[t^1][i][rev[j]]=1;
	}
	For(i,0,n+1){
		memset(sum,0,sizeof(sum));
		t^=1;
		For(j,0,n-i+1) For(k,0,1<<i+j)
			if (f[t][j][k]){
				int x=k>>i,y=k&(1<<i)-1;
				For(w,0,2){
					pii P=trs[j][x][w];
					if (P.fi==-1) continue;
					f[t^1][P.fi][P.se<<(i+1)|y<<1|w]+=f[t][j][k];
				}
				sum[y]+=f[t][j][k];
				f[t][j][k]=0;
			}
		For(x,0,1<<i) if (sum[x]>=k){
			ans=mp(i,x);
			break;
		}
	}
	for (int i=ans.fi-1;~i;i--) printf("%d",ans.se>>i&1);
	puts("");
}
```