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

const int maxn = 22;
int f[2][maxn][(1<<maxn)];
int g[maxn][(1<<maxn)], ans[maxn][(1<<maxn)]; //g[i][j] : i位，状态为j的时候，最前面的一个0的位置
int d[(1<<maxn)]; //d[i] : i有几位（不包括leading 0）
int n;

void init()
{
	for (int i=1;i<=n;i++)
	{
		for (int j=0;j<(1<<i);j++)
		{
			int k = i;
			while ((j>>(k-1))&1) k--;
			g[i][j] = k!=0 ? (i+1-k) : -1;
		}
	}
	for (int i=0;i<(1<<n);i++)
	{
		int digit = 0, ti = i;
		while (ti) digit++, ti >>= 1;
		d[i] = digit;
	}
}

int main()
{
	//freopen("F.in","r",stdin);
	int req;
	cin >> n >> req;
	for (int i=0;i<=n;i++)
	{
		string str;
		cin >> str;
		for (int j=0;j<str.length();j++)
		{
			if (str[j] == '1')
			{
				f[0][i][j] = 1;
			}
		}
	}
	init();
	int cur = 1;
	for (int i=0;i<=n;i++)
	{
		cur = 1 - cur;
		for (int j=1;i+j<=n;j++)
		{
			for (int k=0;k<(1<<(i+j));k++)
			{
				ans[i][k>>j] += f[cur][i+j][k];
				int t = g[j][k & ((1<<j)-1)];
				if (t != -1)
				{
					if (t != j) f[1-cur][i+j-t+1][(k>>j<<(j-t+1)) | (k & ((1<<(j-t))-1))] += f[cur][i+j][k];
					else ans[i+j-t+1][(k>>j<<(j-t+1)) | (k & ((1<<(j-t))-1))] += f[cur][i+j][k];
				}
				if (k & ((1<<j)-1))
				{
					t = d[k & ((1<<j)-1)];
					if (t != 1) f[1-cur][i+t][(k>>j<<t) | (k & ((1<<t)-1))] += f[cur][i+j][k];
					else ans[i+t][(k>>j<<t) | (k & ((1<<t)-1))] += f[cur][i+j][k];
				}
			}
		}
		memset(f[cur],0,sizeof(f[cur]));
	}
	for (int i=n;i>=1;i--)
	{
		for (int j=0;j<(1<<i);j++)
		{
			//cout << i<<" "<<j<<" "<<ans[i][j]<<endl;
			if (ans[i][j] >= req)
			{
				for (int k=i-1;k>=0;k--) 
				{
					cout << ((j >> k) & 1);
				}
				cout << endl;
				return 0;
			}
		}
	}
	return 0;
}

### Prompt

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

const int maxn = 22;
int f[2][maxn][(1<<maxn)];
int g[maxn][(1<<maxn)], ans[maxn][(1<<maxn)]; //g[i][j] : i位，状态为j的时候，最前面的一个0的位置
int d[(1<<maxn)]; //d[i] : i有几位（不包括leading 0）
int n;

void init()
{
	for (int i=1;i<=n;i++)
	{
		for (int j=0;j<(1<<i);j++)
		{
			int k = i;
			while ((j>>(k-1))&1) k--;
			g[i][j] = k!=0 ? (i+1-k) : -1;
		}
	}
	for (int i=0;i<(1<<n);i++)
	{
		int digit = 0, ti = i;
		while (ti) digit++, ti >>= 1;
		d[i] = digit;
	}
}

int main()
{
	//freopen("F.in","r",stdin);
	int req;
	cin >> n >> req;
	for (int i=0;i<=n;i++)
	{
		string str;
		cin >> str;
		for (int j=0;j<str.length();j++)
		{
			if (str[j] == '1')
			{
				f[0][i][j] = 1;
			}
		}
	}
	init();
	int cur = 1;
	for (int i=0;i<=n;i++)
	{
		cur = 1 - cur;
		for (int j=1;i+j<=n;j++)
		{
			for (int k=0;k<(1<<(i+j));k++)
			{
				ans[i][k>>j] += f[cur][i+j][k];
				int t = g[j][k & ((1<<j)-1)];
				if (t != -1)
				{
					if (t != j) f[1-cur][i+j-t+1][(k>>j<<(j-t+1)) | (k & ((1<<(j-t))-1))] += f[cur][i+j][k];
					else ans[i+j-t+1][(k>>j<<(j-t+1)) | (k & ((1<<(j-t))-1))] += f[cur][i+j][k];
				}
				if (k & ((1<<j)-1))
				{
					t = d[k & ((1<<j)-1)];
					if (t != 1) f[1-cur][i+t][(k>>j<<t) | (k & ((1<<t)-1))] += f[cur][i+j][k];
					else ans[i+t][(k>>j<<t) | (k & ((1<<t)-1))] += f[cur][i+j][k];
				}
			}
		}
		memset(f[cur],0,sizeof(f[cur]));
	}
	for (int i=n;i>=1;i--)
	{
		for (int j=0;j<(1<<i);j++)
		{
			//cout << i<<" "<<j<<" "<<ans[i][j]<<endl;
			if (ans[i][j] >= req)
			{
				for (int k=i-1;k>=0;k--) 
				{
					cout << ((j >> k) & 1);
				}
				cout << endl;
				return 0;
			}
		}
	}
	return 0;
}

```