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>
const int N = 21;
int n, m, f[1 << N][N];
int main(){
    scanf("%d%d", &n, &m);
    for(int i = 0;i <= n;++ i)
        for(int j = 0, x;j < (1 << i);++ j){
            for(x = getchar();x != '0' && x != '1';x = getchar());
            f[(1 << i) | j][i] = (x == '1');
        }
    for(int i = n;~i;-- i)
        for(int j = 0;j < (1 << i);++ j)
            for(int k = i;~k;-- k) if(f[(1 << i) | j][k]){
                int t = f[(1 << i) | j][k], S = j >> k, T = j & ((1 << k) - 1);
                if(k) f[(1 << i - k) | S][0] += t;
                if(T){
                    int u = k - 1;
                    while(!((T >> u) & 1)) -- u;
                    f[(1 << i - k + u + 1) | (S << u + 1) | (1 << u) | (T & ((1 << u) - 1))][u] += t;
                }
                if((~T) & ((1 << k) - 1)){
                    int u = k - 1;
                    while((T >> u) & 1) -- u;
                    f[(1 << i - k + 1 + u) | (S << u + 1) | (T & ((1 << u) - 1))][u] += t;
                }
            }
    for(int i = n;~i;-- i)
        for(int j = 0;j < (1 << i);++ j) if(f[(1 << i) | j][0] >= m){
            for(int k = i - 1;~k;-- k) putchar('0' + ((j >> k) & 1));
            return 0;
        }
}