#include <bits/stdc++.h>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int N = 21;
const int MAXS = (1 << N) + 5;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

char g[MAXS];
int n, k, S;
int lg2[MAXS], ch[MAXS][2], dp[N][MAXS];

void init() {
	read(n); read(k);
	S = (1 << (n + 1));
	lg2[0] = -1;
	for (int i = 1; i < S; ++i) {
		lg2[i] = lg2[i >> 1] + 1;
	}
	for (int i = 0; i <= n; ++i) {
		scanf("%s", g);
		int len = (1 << i);
		for (int j = 0; j < len; ++j) {
			dp[i][j ^ len] = (g[j] == '1');
		}
	}
}

void solve() {
	for (int s = 2; s < S; ++s) {
		if ((s >> (lg2[s] - 1)) & 1) {
			ch[s][1] = s ^ (1 << lg2[s]);
			ch[s][0] = ch[ch[s][1]][0];
		} else {
			ch[s][0] = s ^ (3 << (lg2[s] - 1));
			ch[s][1] = ch[ch[s][0]][1];
		}
	}
	for (int i = n; i >= 1; --i) {
		for (int s = (1 << i); s < S; ++s) {
			int x = dp[i][s];
			if (!x) continue;
			int ss = (s >> i), t = (s & ((1 << i) - 1)) ^ (1 << i);
			dp[0][ss] += x;
			if (ch[t][0]) {
				int v = ch[t][0];
				dp[lg2[v]][(((ss << 1) ^ 1) << lg2[v]) ^ v] += x;
			}
			if (ch[t][1]) {
				int v = ch[t][1];
				dp[lg2[v]][((ss << 1) << lg2[v]) ^ v] += x;
			}
		}
	}
	int ans = 1;
	for (int s = S - 1; s > 1; --s) {
		if (dp[0][s] >= k && lg2[s] >= lg2[ans]) {
			ans = s;
		}
	}
	int p = n;
	while (!((ans >> p) & 1)) {
		--p;
	}
	while (p--) {
		putchar(((ans >> p) & 1) + '0');
	}
	puts("");
}

int main() {
	init();
	solve();
	return 0;
}

### Prompt

Please create a solution in cpp 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>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int N = 21;
const int MAXS = (1 << N) + 5;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

char g[MAXS];
int n, k, S;
int lg2[MAXS], ch[MAXS][2], dp[N][MAXS];

void init() {
	read(n); read(k);
	S = (1 << (n + 1));
	lg2[0] = -1;
	for (int i = 1; i < S; ++i) {
		lg2[i] = lg2[i >> 1] + 1;
	}
	for (int i = 0; i <= n; ++i) {
		scanf("%s", g);
		int len = (1 << i);
		for (int j = 0; j < len; ++j) {
			dp[i][j ^ len] = (g[j] == '1');
		}
	}
}

void solve() {
	for (int s = 2; s < S; ++s) {
		if ((s >> (lg2[s] - 1)) & 1) {
			ch[s][1] = s ^ (1 << lg2[s]);
			ch[s][0] = ch[ch[s][1]][0];
		} else {
			ch[s][0] = s ^ (3 << (lg2[s] - 1));
			ch[s][1] = ch[ch[s][0]][1];
		}
	}
	for (int i = n; i >= 1; --i) {
		for (int s = (1 << i); s < S; ++s) {
			int x = dp[i][s];
			if (!x) continue;
			int ss = (s >> i), t = (s & ((1 << i) - 1)) ^ (1 << i);
			dp[0][ss] += x;
			if (ch[t][0]) {
				int v = ch[t][0];
				dp[lg2[v]][(((ss << 1) ^ 1) << lg2[v]) ^ v] += x;
			}
			if (ch[t][1]) {
				int v = ch[t][1];
				dp[lg2[v]][((ss << 1) << lg2[v]) ^ v] += x;
			}
		}
	}
	int ans = 1;
	for (int s = S - 1; s > 1; --s) {
		if (dp[0][s] >= k && lg2[s] >= lg2[ans]) {
			ans = s;
		}
	}
	int p = n;
	while (!((ans >> p) & 1)) {
		--p;
	}
	while (p--) {
		putchar(((ans >> p) & 1) + '0');
	}
	puts("");
}

int main() {
	init();
	solve();
	return 0;
}

```