#include <bits/stdc++.h>

using namespace std;

namespace jumpmelon
{
	const int MAXN = 20;
	char X[(1 << MAXN) + 3];
	int *Ans[MAXN + 1], *F[MAXN + 1][MAXN + 1];
	int *P[MAXN + 1][2];
	
	void work()
	{
		int n, lim;
		scanf("%d%d", &n, &lim);
		for (int i = 0; i <= n; i++)
		{
			Ans[i] = new int[1 << i | 3];
			memset(Ans[i], 0, sizeof(int[1 << i]));
			for (int j = 0; i + j <= n; j++)
			{
				F[i][j] = new int[1 << (i + j) | 3];
				memset(F[i][j], 0, sizeof(int[1 << (i + j)]));
			}
			for (int k = 0; k < 2; k++)
			{
				P[i][k] = new int[1 << i | 3];
				for (int S = 0; S < (1 << i); S++)
				{
					if (i == 0)
						P[i][k][S] = -1;
					else if ((S >> (i - 1) & 1) == k)
						P[i][k][S] = i - 1;
					else
						P[i][k][S] = P[i - 1][k][S & ((1 << (i - 1)) - 1)];
				}
			}
		} 
		for (int j = 0; j <= n; j++)
		{
			scanf("%s", X);
			for (int T = 0; T < (1 << j); T++)
				F[0][j][T] = X[T] - '0';
		}
		for (int i = 0; i <= n; i++)
			for (int j = 0; i + j <= n; j++)
				for (int S = 0; S < (1 << i); S++)
					for (int T = 0; T < (1 << j); T++)
					{
						int v = F[i][j][S << j | T];
						Ans[i][S] += v;
						for (int k = 0; k < 2; k++)
						{
							int m = P[j][k][T], TT = T & ((1 << m) - 1);
							if (m >= 0)
								F[i + 1][m][(S << 1 | k) << m | TT] += v;
						}
					}
		for (int l = n; l >= 0; l--)
			for (int S = 0; S < (1 << l); S++)
				if (Ans[l][S] >= lim)
				{
					for (int i = (l - 1); i >= 0; i--)
						putchar((S >> i & 1) + '0');
					putchar('\n');
					return;
				}
	}
}

int main()
{
	jumpmelon::work();
	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 <bits/stdc++.h>

using namespace std;

namespace jumpmelon
{
	const int MAXN = 20;
	char X[(1 << MAXN) + 3];
	int *Ans[MAXN + 1], *F[MAXN + 1][MAXN + 1];
	int *P[MAXN + 1][2];
	
	void work()
	{
		int n, lim;
		scanf("%d%d", &n, &lim);
		for (int i = 0; i <= n; i++)
		{
			Ans[i] = new int[1 << i | 3];
			memset(Ans[i], 0, sizeof(int[1 << i]));
			for (int j = 0; i + j <= n; j++)
			{
				F[i][j] = new int[1 << (i + j) | 3];
				memset(F[i][j], 0, sizeof(int[1 << (i + j)]));
			}
			for (int k = 0; k < 2; k++)
			{
				P[i][k] = new int[1 << i | 3];
				for (int S = 0; S < (1 << i); S++)
				{
					if (i == 0)
						P[i][k][S] = -1;
					else if ((S >> (i - 1) & 1) == k)
						P[i][k][S] = i - 1;
					else
						P[i][k][S] = P[i - 1][k][S & ((1 << (i - 1)) - 1)];
				}
			}
		} 
		for (int j = 0; j <= n; j++)
		{
			scanf("%s", X);
			for (int T = 0; T < (1 << j); T++)
				F[0][j][T] = X[T] - '0';
		}
		for (int i = 0; i <= n; i++)
			for (int j = 0; i + j <= n; j++)
				for (int S = 0; S < (1 << i); S++)
					for (int T = 0; T < (1 << j); T++)
					{
						int v = F[i][j][S << j | T];
						Ans[i][S] += v;
						for (int k = 0; k < 2; k++)
						{
							int m = P[j][k][T], TT = T & ((1 << m) - 1);
							if (m >= 0)
								F[i + 1][m][(S << 1 | k) << m | TT] += v;
						}
					}
		for (int l = n; l >= 0; l--)
			for (int S = 0; S < (1 << l); S++)
				if (Ans[l][S] >= lim)
				{
					for (int i = (l - 1); i >= 0; i--)
						putchar((S >> i & 1) + '0');
					putchar('\n');
					return;
				}
	}
}

int main()
{
	jumpmelon::work();
	return 0;
}
```