#include <bits/stdc++.h>
using namespace std;
const int maxn = 20;
int n, f[maxn + 1][(1 << maxn) + 10], K;
int g[maxn + 1][(1 << maxn) + 10];
char s[(1 << maxn) + 10];
int sum[(1 << maxn) + 10], h[(1 << maxn) + 10];
int ansl, ans;

int ctz(int x) {
	return x ? h[x] : 32;
}

int main() {
	for (int i = 1; i < 1 << maxn; ++i) h[i] = i & 1 ? 0 : h[i >> 1] + 1;
	scanf("%d%d", &n, &K);
	for (int i = 0; i <= n; ++i) {
		scanf("%s", s);
		for (int j = 0; j < 1 << i; ++j)
			if (s[j] == '1') ++f[i][j];
	}
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j <= n; ++j)
			for (int k = 0; k < 1 << j; ++k) g[j][k] = 0;
		for (int j = i + 1; j <= n; ++j)
			for (int k = 0; k < 1 << j; ++k) {
				int d = ctz(k >> i);
				if (d < j - i)
					g[j - d][((k >> (i + d)) << i) | (k & ((1 << i) - 1))] += f[j][k];
				d = ctz((k >> i) ^ ((1 << (j - i)) - 1));
				if (d < j - i)
					g[j - d][((k >> (i + d)) << i) | (k & ((1 << i) - 1))] += f[j][k];
			}
		for (int j = 0; j <= n; ++j)
			for (int k = 0; k < 1 << j; ++k) f[j][k] = g[j][k];
		for (int j = 0; j < (1 << (i + 1)); ++j) sum[j] = 0;
		for (int j = i + 1; j <= n; ++j)
			for (int k = 0; k < (1 << j); ++k)
				sum[k & ((1 << (i + 1)) - 1)] += f[j][k];
		for (int j = (1 << (i + 1)) - 1; j >= 0; --j)
			if (sum[j] >= K) {
				ansl = i + 1; ans = j;
			}
	}
	for (int i = ansl - 1; i >= 0; --i)
		printf("%d", ans >> i & 1);
}
### 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;
const int maxn = 20;
int n, f[maxn + 1][(1 << maxn) + 10], K;
int g[maxn + 1][(1 << maxn) + 10];
char s[(1 << maxn) + 10];
int sum[(1 << maxn) + 10], h[(1 << maxn) + 10];
int ansl, ans;

int ctz(int x) {
	return x ? h[x] : 32;
}

int main() {
	for (int i = 1; i < 1 << maxn; ++i) h[i] = i & 1 ? 0 : h[i >> 1] + 1;
	scanf("%d%d", &n, &K);
	for (int i = 0; i <= n; ++i) {
		scanf("%s", s);
		for (int j = 0; j < 1 << i; ++j)
			if (s[j] == '1') ++f[i][j];
	}
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j <= n; ++j)
			for (int k = 0; k < 1 << j; ++k) g[j][k] = 0;
		for (int j = i + 1; j <= n; ++j)
			for (int k = 0; k < 1 << j; ++k) {
				int d = ctz(k >> i);
				if (d < j - i)
					g[j - d][((k >> (i + d)) << i) | (k & ((1 << i) - 1))] += f[j][k];
				d = ctz((k >> i) ^ ((1 << (j - i)) - 1));
				if (d < j - i)
					g[j - d][((k >> (i + d)) << i) | (k & ((1 << i) - 1))] += f[j][k];
			}
		for (int j = 0; j <= n; ++j)
			for (int k = 0; k < 1 << j; ++k) f[j][k] = g[j][k];
		for (int j = 0; j < (1 << (i + 1)); ++j) sum[j] = 0;
		for (int j = i + 1; j <= n; ++j)
			for (int k = 0; k < (1 << j); ++k)
				sum[k & ((1 << (i + 1)) - 1)] += f[j][k];
		for (int j = (1 << (i + 1)) - 1; j >= 0; --j)
			if (sum[j] >= K) {
				ansl = i + 1; ans = j;
			}
	}
	for (int i = ansl - 1; i >= 0; --i)
		printf("%d", ans >> i & 1);
}
```