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 <cstdio>
#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <map>
#include <set>

#define MIN_INT -2147483648
#define MAX_INT 2147483647
#define MIN_LONG -9223372036854775808L
#define MAX_LONG 9223372036854775807L
#define PI 3.141592653589793238462643383279502884L

#define long long long int

using std::vector;
using std::map;
using std::set;
using std::string;
using std::pair;
using std::cin;
using std::cout;
using std::cerr;

// @author: pashka

int main() {
    std::ios::sync_with_stdio(false);

    int n, K;
    cin >> n >> K;
    vector<vector<vector<int>>> d(n + 1);
    for (int i = 0; i <= n; i++) {
        d[i].assign(1 << i, vector<int>(i + 1));
        string s;
        cin >> s;
        for (int j = 0; j < (1 << i); j++) {
            if (s[j] == '1') d[i][j][0]++;
        }
    }
    vector<vector<int>> r(n + 1);
    for (int i = 0; i <= n; i++) {
        r[i].assign(1 << i, 0);
    }

    for (int i = n; i > 0; i--) {
        for (int x = 0; x < (1 << i); x++) {
            for (int k = 0; k <= i; k++) {
                int xx = (x & ((1 << k) - 1));
                r[k][xx] += d[i][x][k];
                for (int c = 0; c <= 1; c++) {
                    int kk = k;
                    while (kk < i && ((x >> kk) & 1) != c) kk++;
                    if (kk < i) {
                        xx = x >> (kk - k);
                        xx &= ~((1 << k) - 1);
                        xx |= (x & ((1 << k) - 1));
                        d[i - (kk - k)][xx][k + 1] += d[i][x][k];
                    }
                }
            }
        }
    }

    for (int i = n; i >= 0; i--) {
        for (int x = 0; x < (1 << i); x++) {
            if (r[i][x] >= K) {
                for (int k = i - 1; k >= 0; k--) {
                    cout << ((x >> k) & 1);
                }
                return 0;
            }
        }
    }

    return 0;
}