#include <algorithm>
#include <cstdio>
using namespace std;
int n, k, dp[1 << 21][21], f0[1 << 21], f1[1 << 21];
int main() {
	// freopen("input", "r", stdin);
	scanf("%d%d", &n, &k);
	for(int i = 0; i <= n; i ++) {
		static char s[1 << 21]; scanf("%s", s);
		for(int j = 0; j < (1 << i); j ++) if(s[j] == '1') dp[j | (1 << i)][i] = 1;
	}
	for(int i = 1; i < (1 << (n + 1)); i ++) {
		bool t = 0; f0[i] = f1[i] = -1;
		for(int j = n; j >= 0; j --) {
			bool c = i >> j & 1;
			if(!t) t |= c;
			else {
				if(c && -1 == f1[i]) f1[i] = j;
				if(!c && -1 == f0[i]) f0[i] = j; 
			}
		}
	}
	int ans = 0, len = -1;
	for(int i = (1 << (n + 1)) - 1; i >= 1; i --) {
		int l = 0;
		for(int j = n; j >= 0; j --) if(i >> j & 1) { l = j; break ; }
		for(int j = l; j >= 0; j --) if(dp[i][j]) {
			int t = i & ((1 << j) - 1), s = (i ^ (1 << l)) >> j;
			if(j) dp[s | (1 << (l - j))][0] += dp[i][j];
			int p0 = f0[t | (1 << j)], p1 = f1[t | (1 << j)];
			if(~ p0) dp[(s << (p0 + 1)) | (t & ((1 << p0) - 1)) | (1 << (l - j + p0 + 1))][p0] += dp[i][j];
			if(~ p1) dp[(s << (p1 + 1)) | (1 << p1) | (t & ((1 << p1) - 1)) | (1 << (l - j + p1 + 1))][p1] += dp[i][j];
		}
		if(dp[i][0] >= k && (l > len || (l == len && ans > i))) ans = i, len = l;
	}
	for(int i = n; i >= 0; i --) if(ans >> i & 1) {
		for(int j = i - 1; j >= 0; j --) {
			putchar('0' + (ans >> j & 1));
		}
		break ;
	}
	putchar('\n');
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <algorithm>
#include <cstdio>
using namespace std;
int n, k, dp[1 << 21][21], f0[1 << 21], f1[1 << 21];
int main() {
	// freopen("input", "r", stdin);
	scanf("%d%d", &n, &k);
	for(int i = 0; i <= n; i ++) {
		static char s[1 << 21]; scanf("%s", s);
		for(int j = 0; j < (1 << i); j ++) if(s[j] == '1') dp[j | (1 << i)][i] = 1;
	}
	for(int i = 1; i < (1 << (n + 1)); i ++) {
		bool t = 0; f0[i] = f1[i] = -1;
		for(int j = n; j >= 0; j --) {
			bool c = i >> j & 1;
			if(!t) t |= c;
			else {
				if(c && -1 == f1[i]) f1[i] = j;
				if(!c && -1 == f0[i]) f0[i] = j; 
			}
		}
	}
	int ans = 0, len = -1;
	for(int i = (1 << (n + 1)) - 1; i >= 1; i --) {
		int l = 0;
		for(int j = n; j >= 0; j --) if(i >> j & 1) { l = j; break ; }
		for(int j = l; j >= 0; j --) if(dp[i][j]) {
			int t = i & ((1 << j) - 1), s = (i ^ (1 << l)) >> j;
			if(j) dp[s | (1 << (l - j))][0] += dp[i][j];
			int p0 = f0[t | (1 << j)], p1 = f1[t | (1 << j)];
			if(~ p0) dp[(s << (p0 + 1)) | (t & ((1 << p0) - 1)) | (1 << (l - j + p0 + 1))][p0] += dp[i][j];
			if(~ p1) dp[(s << (p1 + 1)) | (1 << p1) | (t & ((1 << p1) - 1)) | (1 << (l - j + p1 + 1))][p1] += dp[i][j];
		}
		if(dp[i][0] >= k && (l > len || (l == len && ans > i))) ans = i, len = l;
	}
	for(int i = n; i >= 0; i --) if(ans >> i & 1) {
		for(int j = i - 1; j >= 0; j --) {
			putchar('0' + (ans >> j & 1));
		}
		break ;
	}
	putchar('\n');
	return 0;
}
```