#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <fstream>
#include <iomanip>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int MAXN = 1 << 21;

int n,K;
int len[MAXN];
int top[MAXN];
int nxt[MAXN];
int sum[MAXN];
int f[21][MAXN];

char s[MAXN];

int getid(int d,int i,int j)
{
	return (i - (1 << d)) << (n - d + 1) | j;
}

int main()
{
	scanf("%d%d",&n,&K);
	for (int i = 0;i <= n;i++)
	{
		scanf("%s",s);
		for (int j = 0;j < (1 << i);j++)
			sum[j + (1 << i)] = s[j] - '0';
	}
	for (int i = (1 << n) - 1;i >= 1;i--)
		sum[i] += sum[i << 1] + sum[i << 1 | 1];
	for (int i = 1;i < (1 << (n + 1));i++)
	{
		for (int j = n;j >= 0;j--)
			if (i >> j & 1)
			{
				if (j)
				{
					len[i] = j;
					top[i] = (i >> (j - 1) & 1);
					nxt[i] = (i - (1 << j)) | (1 << (j - 1));
				}
				break;
			}
	}
	for (int d = n;d >= 0;d--)
		for (int i = (1 << (d + 1)) - 1;i >= (1 << d);i--)
		{
			f[d][getid(d,i,1)] = sum[i];
			for (int j = 2;j < (1 << (n - d + 1));j++)
			{
				f[d][getid(d,i,j)] = f[d + 1][getid(d + 1,i << 1 | top[j],nxt[j])];
				if (len[j] <= n - d - 1)
					f[d][getid(d,i,j)] += f[d + 1][getid(d + 1,i << 1 | (top[j] ^ 1),j)];
			}
		}
	int MX = -1,ans = 0;
	for (int i = 1;i < (1 << (n + 1));i++)
		if (f[0][i] >= K && len[i] > MX)
		{
			MX = len[i];
			ans = i;
		}
	for (int i = len[ans] - 1;i >= 0;i--)
		printf("%d",ans >> i & 1);
	printf("\n");
    return 0;
}
### Prompt

In cpp, your task is to solve 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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <fstream>
#include <iomanip>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int MAXN = 1 << 21;

int n,K;
int len[MAXN];
int top[MAXN];
int nxt[MAXN];
int sum[MAXN];
int f[21][MAXN];

char s[MAXN];

int getid(int d,int i,int j)
{
	return (i - (1 << d)) << (n - d + 1) | j;
}

int main()
{
	scanf("%d%d",&n,&K);
	for (int i = 0;i <= n;i++)
	{
		scanf("%s",s);
		for (int j = 0;j < (1 << i);j++)
			sum[j + (1 << i)] = s[j] - '0';
	}
	for (int i = (1 << n) - 1;i >= 1;i--)
		sum[i] += sum[i << 1] + sum[i << 1 | 1];
	for (int i = 1;i < (1 << (n + 1));i++)
	{
		for (int j = n;j >= 0;j--)
			if (i >> j & 1)
			{
				if (j)
				{
					len[i] = j;
					top[i] = (i >> (j - 1) & 1);
					nxt[i] = (i - (1 << j)) | (1 << (j - 1));
				}
				break;
			}
	}
	for (int d = n;d >= 0;d--)
		for (int i = (1 << (d + 1)) - 1;i >= (1 << d);i--)
		{
			f[d][getid(d,i,1)] = sum[i];
			for (int j = 2;j < (1 << (n - d + 1));j++)
			{
				f[d][getid(d,i,j)] = f[d + 1][getid(d + 1,i << 1 | top[j],nxt[j])];
				if (len[j] <= n - d - 1)
					f[d][getid(d,i,j)] += f[d + 1][getid(d + 1,i << 1 | (top[j] ^ 1),j)];
			}
		}
	int MX = -1,ans = 0;
	for (int i = 1;i < (1 << (n + 1));i++)
		if (f[0][i] >= K && len[i] > MX)
		{
			MX = len[i];
			ans = i;
		}
	for (int i = len[ans] - 1;i >= 0;i--)
		printf("%d",ans >> i & 1);
	printf("\n");
    return 0;
}
```