#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,f[2200005][22],nxt[2200005][4];
char s[1200005];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(int j=0;j<(1<<i);j++)
			if(s[j]=='1') f[(1<<i)+j][i]=1;
	}
	//nxt[1][0]=nxt[1][1]=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<(1<<i);j++)
		{
			int nw=(1<<i)+j;
			if(j&(1<<(i-1)))
			{
				nxt[nw][1]=nw-(1<<i);
				nxt[nw][0]=nxt[nw-(1<<i)][0];
				nxt[nw][2]=nxt[nw-(1<<i)][2];
				nxt[nw][3]=i-1;
			}
			else
			{
				nxt[nw][0]=nw-(1<<(i-1));
				nxt[nw][1]=nxt[nw-(1<<(i-1))][1];
				nxt[nw][3]=nxt[nw-(1<<(i-1))][3];
				nxt[nw][2]=i-1;
			}
		//	printf("i=%d,j=%d,nw=%d,nxt=%d,%d,%d,%d\n"
		//	,i,j,nw,nxt[nw][0],nxt[nw][1],nxt[nw][2],nxt[nw][3]);
		}
	}
	for(int i=n;i>0;i--)
		for(int j=i;j<=n;j++)
		{
			for(int k=0;k<(1<<j);k++)
			{
				int nw=(1<<j)+k,t=(k&((1<<i)-1))+(1<<i),tnw=(nw>>i)<<i;
				int len0=nxt[t][2],len1=nxt[t][3],nw0=nxt[t][0]-(1<<len0),nw1=nxt[t][1];
				f[tnw>>i][0]+=f[nw][i];
				if(nxt[t][0]) f[(tnw>>(i-len0-1))+nw0][len0]+=f[nw][i];
				if(nxt[t][1]) f[(tnw>>(i-len1-1))+nw1][len1]+=f[nw][i];
				//if(f[nw][i])printf("i=%d,j=%d,k=%d,nw=%d,t=%d,len=%d,%d,nw=%d,%d,f=%d,nxtf=(%d,%d),(%d,%d)\n"
			//	,i,j,k,nw,t,len0,len1,nw0,nw1,f[nw][i],(tnw>>(i-len0-1))+nw0,len0,(tnw>>(i-len1-1))+nw1,len1);
			}
		}
	for(int i=n;i>=0;i--)
		for(int j=0;j<(1<<i);j++)
		{
			if(f[(1<<i)+j][0]>=m)
			{
			//	printf("i=%d,j=%d\n",i,j);
				for(int k=i-1;k>=0;k--)
					printf("%d",(j>>k)&1);
				return 0;
			}
		}
	return 0;
}
### Prompt

Create a solution in Cpp for 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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,f[2200005][22],nxt[2200005][4];
char s[1200005];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(int j=0;j<(1<<i);j++)
			if(s[j]=='1') f[(1<<i)+j][i]=1;
	}
	//nxt[1][0]=nxt[1][1]=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<(1<<i);j++)
		{
			int nw=(1<<i)+j;
			if(j&(1<<(i-1)))
			{
				nxt[nw][1]=nw-(1<<i);
				nxt[nw][0]=nxt[nw-(1<<i)][0];
				nxt[nw][2]=nxt[nw-(1<<i)][2];
				nxt[nw][3]=i-1;
			}
			else
			{
				nxt[nw][0]=nw-(1<<(i-1));
				nxt[nw][1]=nxt[nw-(1<<(i-1))][1];
				nxt[nw][3]=nxt[nw-(1<<(i-1))][3];
				nxt[nw][2]=i-1;
			}
		//	printf("i=%d,j=%d,nw=%d,nxt=%d,%d,%d,%d\n"
		//	,i,j,nw,nxt[nw][0],nxt[nw][1],nxt[nw][2],nxt[nw][3]);
		}
	}
	for(int i=n;i>0;i--)
		for(int j=i;j<=n;j++)
		{
			for(int k=0;k<(1<<j);k++)
			{
				int nw=(1<<j)+k,t=(k&((1<<i)-1))+(1<<i),tnw=(nw>>i)<<i;
				int len0=nxt[t][2],len1=nxt[t][3],nw0=nxt[t][0]-(1<<len0),nw1=nxt[t][1];
				f[tnw>>i][0]+=f[nw][i];
				if(nxt[t][0]) f[(tnw>>(i-len0-1))+nw0][len0]+=f[nw][i];
				if(nxt[t][1]) f[(tnw>>(i-len1-1))+nw1][len1]+=f[nw][i];
				//if(f[nw][i])printf("i=%d,j=%d,k=%d,nw=%d,t=%d,len=%d,%d,nw=%d,%d,f=%d,nxtf=(%d,%d),(%d,%d)\n"
			//	,i,j,k,nw,t,len0,len1,nw0,nw1,f[nw][i],(tnw>>(i-len0-1))+nw0,len0,(tnw>>(i-len1-1))+nw1,len1);
			}
		}
	for(int i=n;i>=0;i--)
		for(int j=0;j<(1<<i);j++)
		{
			if(f[(1<<i)+j][0]>=m)
			{
			//	printf("i=%d,j=%d\n",i,j);
				for(int k=i-1;k>=0;k--)
					printf("%d",(j>>k)&1);
				return 0;
			}
		}
	return 0;
}
```