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 std::cerr;
using std::endl;

const int N = 21;

int n, K, ans = 1, h[(1 << N) + 5], ch[(1 << N) + 5][2], f[N][(1 << N) + 5];

int main() {
  scanf("%d %d", &n, &K);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j < 1 << i; ++j)
      scanf("%1d", &f[i][(1 << i) | j]);
  h[0] = -1;
  for (int s = 1; s < 2 << n; ++s)
    h[s] = h[s >> 1] + 1;
  for (int s = 2; s < 2 << n; ++s) {
    if (s & (1 << (h[s] - 1))) {
      ch[s][1] = s ^ (1 << h[s]);
      ch[s][0] = ch[ch[s][1]][0];
    } else {
      ch[s][0] = s ^ (3 << (h[s] - 1));
      ch[s][1] = ch[ch[s][0]][1];
    }
  }
  for (int i = n; i; --i) {
    for (int v = 1 << i; v < 2 << n; ++v) {
      int s = v >> i, t = (v & ((1 << i) - 1)) | (1 << i);
      if (ch[t][0])
        f[h[ch[t][0]]][((s << 1 ^ 1) << h[ch[t][0]]) ^ ch[t][0]] += f[i][v];
      if (ch[t][1])
        f[h[ch[t][1]]][((s << 1 ^ 0) << h[ch[t][1]]) ^ ch[t][1]] += f[i][v];
      f[0][s] += f[i][v];
    }
  }
  for (int s = 2; s < 2 << n; ++s)
    if (f[0][s] >= K && h[s] > h[ans])
      ans = s;
  int p = n;
  while (!(ans >> p) & 1)
    --p;
  while (p--)
    printf("%d", (ans >> p) & 1);
  putchar('\n');
  return 0;
}