#include<bits/stdc++.h>
using namespace std;

char p1[42000010];
int n,k,bg[21][1<<20|3],cnt,p2[42000010],h1[1<<20|3],h0[1<<20|3],an,ans,f[42000010];
char s[21][1<<20|3];

int main()
{
	h1[0]=-1,h1[1]=0;
	for (int i=2; i<(1<<20); i++) h1[i]=h1[i>>1]+1;
	h0[0]=h0[1]=-1;
	for (int i=2; i<(1<<20); i++)
	{
		h0[i]=-1;
		for (int j=h1[i]; j>=0; j--)
			if (!(i>>j&1)) {h0[i]=j; break;}
	}
	scanf("%d%d",&n,&k),cnt=0;
	for (int i=0; i<=n; i++) scanf("%s",s[i]);
	for (int i=0; i<=n; i++)
		for (int j=0; j<(1<<i); j++)
		{
			bg[i][j]=(++cnt);
			for (int k=cnt; k<=cnt+i; k++) p1[k]=i,p2[k]=j;
			cnt+=i;
		}
	an=0;
	for (int i=cnt; i; i--) 
		if (i!=bg[p1[i]][p2[i]])
		{
			if (i==bg[p1[i]][p2[i]]+p1[i]&&s[p1[i]][p2[i]]=='1') f[i]++;
			int len=i-bg[p1[i]][p2[i]];
			f[bg[p1[i]-len][p2[i]>>len]]+=f[i];
			int p=h1[p2[i]&((1<<len)-1)];
			if (p!=-1) f[bg[p1[i]-(len-1-p)][((p2[i]>>len)<<p+1)|(p2[i]&((1<<p)-1))|(1<<p)]+p]+=f[i];
			if (p!=len-1) p=len-1; else p=h0[p2[i]&((1<<len)-1)];
			if (p!=-1) f[bg[p1[i]-(len-1-p)][((p2[i]>>len)<<p+1)|(p2[i]&((1<<p)-1))]+p]+=f[i];
		} else
		{
			if (f[i]<k) continue;
			if (p1[i]>an) an=p1[i],ans=p2[i]; else
			if (p1[i]==an&&ans>p2[i]) ans=p2[i]; 
		}
	for (int i=an-1; i>=0; i--) putchar((ans>>i&1)+'0');
	return puts(""),0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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;

char p1[42000010];
int n,k,bg[21][1<<20|3],cnt,p2[42000010],h1[1<<20|3],h0[1<<20|3],an,ans,f[42000010];
char s[21][1<<20|3];

int main()
{
	h1[0]=-1,h1[1]=0;
	for (int i=2; i<(1<<20); i++) h1[i]=h1[i>>1]+1;
	h0[0]=h0[1]=-1;
	for (int i=2; i<(1<<20); i++)
	{
		h0[i]=-1;
		for (int j=h1[i]; j>=0; j--)
			if (!(i>>j&1)) {h0[i]=j; break;}
	}
	scanf("%d%d",&n,&k),cnt=0;
	for (int i=0; i<=n; i++) scanf("%s",s[i]);
	for (int i=0; i<=n; i++)
		for (int j=0; j<(1<<i); j++)
		{
			bg[i][j]=(++cnt);
			for (int k=cnt; k<=cnt+i; k++) p1[k]=i,p2[k]=j;
			cnt+=i;
		}
	an=0;
	for (int i=cnt; i; i--) 
		if (i!=bg[p1[i]][p2[i]])
		{
			if (i==bg[p1[i]][p2[i]]+p1[i]&&s[p1[i]][p2[i]]=='1') f[i]++;
			int len=i-bg[p1[i]][p2[i]];
			f[bg[p1[i]-len][p2[i]>>len]]+=f[i];
			int p=h1[p2[i]&((1<<len)-1)];
			if (p!=-1) f[bg[p1[i]-(len-1-p)][((p2[i]>>len)<<p+1)|(p2[i]&((1<<p)-1))|(1<<p)]+p]+=f[i];
			if (p!=len-1) p=len-1; else p=h0[p2[i]&((1<<len)-1)];
			if (p!=-1) f[bg[p1[i]-(len-1-p)][((p2[i]>>len)<<p+1)|(p2[i]&((1<<p)-1))]+p]+=f[i];
		} else
		{
			if (f[i]<k) continue;
			if (p1[i]>an) an=p1[i],ans=p2[i]; else
			if (p1[i]==an&&ans>p2[i]) ans=p2[i]; 
		}
	for (int i=an-1; i>=0; i--) putchar((ans>>i&1)+'0');
	return puts(""),0;
}
```