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 <algorithm>
#include <vector>
constexpr int N = 20;
int n, k;
int dp[1 << (N + 1)][N + 1], to[1 << (N + 1)][2];
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cin >> n >> k;
    for (int len = 0; len <= n; ++len) {
        for (int s = 0; s < (1 << len); ++s) {
            int t = s | 1 << len;
            int i = len - 1;
            while (i >= 0 && (s >> i & 1))
                --i;
            to[t][0] = i;
            i = len - 1;
            while (i >= 0 && (~s >> i & 1))
                --i;
            to[t][1] = i;
        }
    }
    for (int len = 0; len <= n; ++len) {
        for (int s = 0; s < (1 << len); ++s) {
            char c;
            std::cin >> c;
            if (c == '1')
                dp[s | 1 << len][0] = 1;
        }
    }
    for (int len = n; len >= 0; --len) {
        for (int s = 0; s < (1 << len); ++s) {
            for (int p = 0; p < len; ++p) {
                int t = len - p;
                int suf = s & ((1 << t) - 1) | 1 << t, pre = s >> t | 1 << p;
                if (to[suf][0] != -1)
                    dp[pre << (to[suf][0] + 1) | s & ((1 << to[suf][0]) - 1)][p + 1] += dp[s | 1 << len][p];
                if (to[suf][1] != -1)
                    dp[(pre << 1 | 1) << to[suf][1] | s & ((1 << to[suf][1]) - 1)][p + 1] += dp[s | 1 << len][p];
                dp[pre][p] += dp[s | 1 << len][p];
            }
        }
    }
    for (int len = n; len >= 0; --len) {
        for (int s = 0; s < (1 << len); ++s) {
            if (dp[s | 1 << len][len] >= k) {
                for (int i = len - 1; i >= 0; --i)
                    std::cout << (s >> i & 1);
                std::cout << "\n";
                return 0;
            }
        }
    }
}