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 <bits/stdc++.h>
using namespace std;

const int N = 1048576;
int n, K, f[21][N], g[21][N];
char s[N];

int main() {
  scanf("%d%d", &n, &K);
  for (int i = 0; i <= n; i++) {
    scanf("%s", s);
    for (int j = 0; j < 1 << i; j++) {
      f[i][j] = s[j] - '0';
    }
  }
  for (int i = 1; i <= n; i++) {
    g[i][0] = -1;
    for (int j = 1; j < 1 << i - 1; j++) {
      g[i][j] = 32 - __builtin_clz(j);
    }
    for (int j = 1 << i - 1; j < 1 << i; j++) {
      g[i][j] = 32 - __builtin_clz(j ^ (1 << i) - 1);
    }
    g[i][(1 << i) - 1] = -1;
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 1; i + j <= n; j++) {
      for (int k = 0; k < 1 << i + j; k++) {
        f[i][k >> j] += f[i + j][k];
        int r = g[j][k & (1 << j) - 1];
        if (~r) f[i + r][k >> j << r | (k & (1 << r) - 1)] += f[i + j][k];
      }
    }
  }
  int len = 0, ans = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = (1 << i) - 1; ~j; j--) {
      if (f[i][j] >= K) {
        len = i, ans = j;
      }
    }
  }
  for (int j = len - 1; ~j; j--) {
    putchar(48 + (ans >> j & 1));
  }
}