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 "cstring"
#include "cstdio"
using namespace std;
#define MAXN 21
int n , k , dp[1 << 21][21];
char ch[1 << 21];
int main() {
    cin >> n >> k;
    for( int i = 0 ; i <= n ; ++ i ) {
        scanf("%s",ch);
        for( int j = 0 ; j < ( 1 << i ) ; ++ j )
            dp[( 1 << i ) | j][i] = ( ch[j] == '1' );
    }
    for( int i = n ; i >= 0 ; -- i ) {
        for( int j = 0 ; j < ( 1 << i ) ; ++ j ) // sta : ( 1 << i ) | j
            for( int k = i ; k >= 0 ; -- k ) if( dp[( 1 << i ) | j][k] ) {
                int S = ( j >> k ) , T = ( j & ( 1 << k ) - 1 ) , f = dp[( 1 << i ) | j][k];
                if( k ) dp[( 1 << i - k ) | S][0] += f;
                int ls = T & ( 1 << k ) - 1 , re = (~T) & ( 1 << k ) - 1;
                if( T ) {
                    int u = 31 - __builtin_clz( ls );
//                    cout << ls << ' ' << u << endl;
                    dp[( 1 << i - k + u + 1 ) | ( S << u + 1 ) | ( 1 << u ) | ( T & ( 1 << u ) - 1 )][u] += f;
                }
                if( re ) {
                    int u = 31 - __builtin_clz( re );
//                    cout << T << ' ' << u << ' ' << k << endl;
                    dp[( 1 << i - k + u + 1 ) | ( S << u + 1 ) | ( T & ( 1 << u ) - 1)][u] += f;
                }
            }
    }
    for( int i = n ; i >= 0 ; -- i )
        for( int j = 0 ; j < ( 1 << i ) ; ++ j ) if( dp[( 1 << i ) | j][0] >= k ) {
            for( int k = i - 1 ; k >= 0 ; -- k ) printf("%d",( j >> k ) & 1);
            return 0;
        }
}