#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
//#include <bits/stdc++.h>
using namespace std;


typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


string s;

int blen = 0;
int bans = 0;

int dp[21][1 << 20];
int tr[21][1 << 20];

void upd(int len, int ans) {
	if (len < blen)
		return;
	if (len > blen) {
		blen = len, bans = ans;
	}
	else if (len == blen && ans < bans) {
		bans = ans;
	}
}

int n, k;

int main() {
	cin >> n >> k;
	for (int i = 0; i <= n; ++i) {
		cin >> s;
		for (int j = 0; j < (1 << i); ++j) {
			if (s[j] == '1') {
				++dp[i][j];
			}
		}
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 0; j < (1 << i); ++j) {
			int x = (j >> (i - 1)) & 1;
			int ans = 0;
			while (ans < i && ((j >> (i - ans - 1)) & 1) == x)
				++ans;
			if (ans == i)
				tr[i][j] = -1;
			else
				tr[i][j] = ans;
		}
	for (int l = 0; l <= n; ++l) {
		for (int i = l + 1; i <= n; ++i) {
			int msk = (1 << (i - l)) - 1;
			for (int j = 0; j < (1 << i); ++j) {
				dp[l][j >> (i - l)] += dp[i][j];
				int x = tr[i - l][j & msk];
				if (x != -1) {
					dp[i - x][((j >> (i - l)) << (i - l - x)) + (j & ((1 << (i - l - x)) - 1))] += dp[i][j];
				}
			}
		}
		for (int i = 0; i < (1 << l); ++i)
			if (dp[l][i] >= k)
				upd(l, i);
	}
	for (int i = blen - 1; i >= 0; --i) {
		cout << ((bans >> i) & 1);
	}
	cout << "\n";
	return 0;
}



### Prompt

Construct a Cpp code solution to the problem outlined:

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 <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
//#include <bits/stdc++.h>
using namespace std;


typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


string s;

int blen = 0;
int bans = 0;

int dp[21][1 << 20];
int tr[21][1 << 20];

void upd(int len, int ans) {
	if (len < blen)
		return;
	if (len > blen) {
		blen = len, bans = ans;
	}
	else if (len == blen && ans < bans) {
		bans = ans;
	}
}

int n, k;

int main() {
	cin >> n >> k;
	for (int i = 0; i <= n; ++i) {
		cin >> s;
		for (int j = 0; j < (1 << i); ++j) {
			if (s[j] == '1') {
				++dp[i][j];
			}
		}
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 0; j < (1 << i); ++j) {
			int x = (j >> (i - 1)) & 1;
			int ans = 0;
			while (ans < i && ((j >> (i - ans - 1)) & 1) == x)
				++ans;
			if (ans == i)
				tr[i][j] = -1;
			else
				tr[i][j] = ans;
		}
	for (int l = 0; l <= n; ++l) {
		for (int i = l + 1; i <= n; ++i) {
			int msk = (1 << (i - l)) - 1;
			for (int j = 0; j < (1 << i); ++j) {
				dp[l][j >> (i - l)] += dp[i][j];
				int x = tr[i - l][j & msk];
				if (x != -1) {
					dp[i - x][((j >> (i - l)) << (i - l - x)) + (j & ((1 << (i - l - x)) - 1))] += dp[i][j];
				}
			}
		}
		for (int i = 0; i < (1 << l); ++i)
			if (dp[l][i] >= k)
				upd(l, i);
	}
	for (int i = blen - 1; i >= 0; --i) {
		cout << ((bans >> i) & 1);
	}
	cout << "\n";
	return 0;
}



```