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

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

// 入力
int N, K;
int dp[1 << 21][22];
string S[1 << 18];

// 遷移
int LEN[1 << 21], BASE[1 << 21];
pair<int, int> G[1 << 21][22];
bool H[1 << 21][22];

void init() {
	for (int i = 0; i < 21; i++) {
		for (int j = (1 << i); j < (2 << i); j++) LEN[j] = i;
	}
	for (int i = 0; i < (1 << 21); i++) BASE[i] = i - (1 << LEN[i]);

	for (int i = 1; i < (2 << N); i++) {
		int bit[22], base = BASE[i];
		for (int j = 0; j < LEN[i]; j++) {
			if ((base & (1 << j)) == 0) bit[j] = 0;
			else bit[j] = 1;
		}

		// 部分消し
		for (int j = 0; j < LEN[i]; j++) {
			int lims = j;
			for (int k = j; k < LEN[i]; k++) { if (k + 1 == LEN[i] || bit[k] != bit[k + 1]) break; lims = k + 1; }

			int r1 = (base & ((1 << j) - 1));
			int r2 = (base >> (lims + 1));
			int cx = r1 + (r2 << j);
			int cp = j + (LEN[i] - lims - 1);
			G[i][j] = make_pair((1 << cp) + cx, min(cp, j + 1));
			if (lims + 1 == LEN[i]) H[i][j] = true;
		}
	}
}

int main() {
	cin >> N >> K;
	for (int i = 0; i <= N; i++) cin >> S[i];
	for (int i = 0; i <= N; i++) {
		for (int j = 0; j < S[i].size(); j++) {
			if (S[i][j] == '1') dp[(1 << i) + j][0] = 1;
		}
	}
	/*string T; cin >> T;
	int r = 0;
	for (int i = 0; i < T.size(); i++) {
		if (T[T.size() - 1 - i] == '1') r += (1 << i);
	}
	dp[(1 << T.size()) + r][0] = 1;*/

	init();

	for (int i = (2 << N) - 1; i >= 1; i--) {
		int cv[22];
		for (int j = 0; j < N; j++) cv[j] = (1 << j) + (BASE[i] & ((1 << j) - 1));
		for (int j = 0; j < LEN[i]; j++) {
			if (dp[i][j] == 0) continue;
			for (int k = j; k < LEN[i]; k++) dp[G[i][k].first][G[i][k].second] += dp[i][j];
			for (int k = j; k < LEN[i]; k++) { if (H[i][k] == false) dp[cv[k]][k] += dp[i][j]; }
		}
	}

	for (int i = N; i >= 0; i--) {
		for (int j = (1 << i); j < (2 << i); j++) {
			int ret = 0;
			for (int k = 0; k <= N; k++) ret += dp[j][k];
			if (ret < K) continue;

			string str = "";
			for (int k = i - 1; k >= 0; k--) {
				if (((j - (1 << i)) / (1 << k)) % 2 == 1) str += "1";
				else str += "0";
			}
			cout << str << endl;
			return 0;
		}
	}
	cout << "-1" << endl;
	return 0;
}