#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std;

namespace zyt
{
	const int N = 21, S = 1 << N | 13;
	int n, dp[N][S], ans[S], first[S];
	bool check(const int a, const int p)
	{
		return a & (1 << p);
	}
	void init()
	{
		first[0] = -1;
		for (int i = 1; i <= N; i++)
			for (int j = 1 << (i - 1); j < (1 << i); j++)
				first[j] = i - 1;
	}
	int work()
	{
		init();
		int k;
		scanf("%d%d", &n, &k);
		for (int i = 0; i <= n; i++)
		{
			static char str[S];
			scanf("%s", str);
			for (int j = 0; j < (1 << i); j++)
				dp[i][j | (1 << i)] = (str[j] == '1');
		}
		for (int i = n; i >= 0; i--)
			for (int j = 1; j < (1 << (n + 1)); j++)
			{
				if (!dp[i][j])
					continue;
				int p = first[j];
				int s1 = (j ^ (1 << p)) >> i, s2 = (j ^ (1 << p)) & ((1 << i) - 1);
				if ((p = first[~s2 & ((1 << i) - 1)]) != -1)
					dp[p][((j >> i << 1) << p) | (s2 & ((1 << p) - 1))] += dp[i][j];
				if ((p = first[s2]) != -1)
					dp[p][(((j >> i << 1) | 1) << p) | (s2 & ((1 << p) - 1))] += dp[i][j];
				ans[j >> i] += dp[i][j];
			}
		int res = 0;
		for (int i = 1; i < (1 << (n + 1)); i++)
			if (ans[i] >= k && first[i] > first[res])
				res = i;
		for (int i = first[res] - 1; i >= 0; i--)
			putchar(check(res, i) ? '1' : '0');
		return 0;
	}
}
int main()
{
	return zyt::work();
}
### Prompt

Your challenge is to write a cpp solution 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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std;

namespace zyt
{
	const int N = 21, S = 1 << N | 13;
	int n, dp[N][S], ans[S], first[S];
	bool check(const int a, const int p)
	{
		return a & (1 << p);
	}
	void init()
	{
		first[0] = -1;
		for (int i = 1; i <= N; i++)
			for (int j = 1 << (i - 1); j < (1 << i); j++)
				first[j] = i - 1;
	}
	int work()
	{
		init();
		int k;
		scanf("%d%d", &n, &k);
		for (int i = 0; i <= n; i++)
		{
			static char str[S];
			scanf("%s", str);
			for (int j = 0; j < (1 << i); j++)
				dp[i][j | (1 << i)] = (str[j] == '1');
		}
		for (int i = n; i >= 0; i--)
			for (int j = 1; j < (1 << (n + 1)); j++)
			{
				if (!dp[i][j])
					continue;
				int p = first[j];
				int s1 = (j ^ (1 << p)) >> i, s2 = (j ^ (1 << p)) & ((1 << i) - 1);
				if ((p = first[~s2 & ((1 << i) - 1)]) != -1)
					dp[p][((j >> i << 1) << p) | (s2 & ((1 << p) - 1))] += dp[i][j];
				if ((p = first[s2]) != -1)
					dp[p][(((j >> i << 1) | 1) << p) | (s2 & ((1 << p) - 1))] += dp[i][j];
				ans[j >> i] += dp[i][j];
			}
		int res = 0;
		for (int i = 1; i < (1 << (n + 1)); i++)
			if (ans[i] >= k && first[i] > first[res])
				res = i;
		for (int i = first[res] - 1; i >= 0; i--)
			putchar(check(res, i) ? '1' : '0');
		return 0;
	}
}
int main()
{
	return zyt::work();
}
```