#include <bits/stdc++.h>

using namespace std;

int n, m;
int f[1 << 21];
int p[1 << 21];
bool e[21][1 << 21];
bool vis[1 << 21];
string s[21];

int calc(int x, int y) {
	int tmp = x & ((1 << y) - 1);
	x >>= (y + 1), x <<= y, x += tmp;
	return x;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for (int i = 0; i <= n; ++i) {
		cin >> s[i];
		for (int j = 0; j < (1 << i); ++j) {
			if (s[i][j] == '1') {
				f[j + (1 << i)] = 1;
				// for (int k = 0; k < i; ++k) cout << (j >> k & 1); cout << '\n';
			}
		}
	}
	int lim = (1 << (n + 1)) - 1;
	for (int i = 1; i <= lim; ++i) {
		for (int j = 0; j <= n; ++j) {
			if (i >> j & 1) p[i] = j;
		}
		for (int j = 0; j < p[i]; ++j) {
			int k = calc(i, j);
			e[j][i] = !vis[k], vis[k] = 1;
		}
		for (int j = 0; j < p[i]; ++j) {
			int k = calc(i, j);
			vis[k] = 0;			
		}
	}
	for (int i = 0; i <= n; ++i) {
		for (int j = lim; j >= 1; --j) {
			if (!e[i][j]) continue;
			int k = calc(j, i);
			f[k] += f[j];
		}
	}
	int res = 0;
	for (int i = 1; i <= lim; ++i) {
		if (f[i] >= m) {
			if (!res || p[i] > p[res]) res = i;
		}
	}
	int p = 0;
	for (int i = 0; i <= n; ++i) {
		if (res >> i & 1) p = i;
	}
	for (int i = p - 1; i >= 0; --i) {
		cout << (res >> i & 1);
	}
}
### Prompt

Please formulate 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;

int n, m;
int f[1 << 21];
int p[1 << 21];
bool e[21][1 << 21];
bool vis[1 << 21];
string s[21];

int calc(int x, int y) {
	int tmp = x & ((1 << y) - 1);
	x >>= (y + 1), x <<= y, x += tmp;
	return x;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for (int i = 0; i <= n; ++i) {
		cin >> s[i];
		for (int j = 0; j < (1 << i); ++j) {
			if (s[i][j] == '1') {
				f[j + (1 << i)] = 1;
				// for (int k = 0; k < i; ++k) cout << (j >> k & 1); cout << '\n';
			}
		}
	}
	int lim = (1 << (n + 1)) - 1;
	for (int i = 1; i <= lim; ++i) {
		for (int j = 0; j <= n; ++j) {
			if (i >> j & 1) p[i] = j;
		}
		for (int j = 0; j < p[i]; ++j) {
			int k = calc(i, j);
			e[j][i] = !vis[k], vis[k] = 1;
		}
		for (int j = 0; j < p[i]; ++j) {
			int k = calc(i, j);
			vis[k] = 0;			
		}
	}
	for (int i = 0; i <= n; ++i) {
		for (int j = lim; j >= 1; --j) {
			if (!e[i][j]) continue;
			int k = calc(j, i);
			f[k] += f[j];
		}
	}
	int res = 0;
	for (int i = 1; i <= lim; ++i) {
		if (f[i] >= m) {
			if (!res || p[i] > p[res]) res = i;
		}
	}
	int p = 0;
	for (int i = 0; i <= n; ++i) {
		if (res >> i & 1) p = i;
	}
	for (int i = p - 1; i >= 0; --i) {
		cout << (res >> i & 1);
	}
}
```