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;
#define FOR(i, a, b)for(int i = (a);i <= (b); i++)
#define ROF(i, a, b)for(int i = (a);i >= (b); i--)
const int N = 22;
int f[N][1<<N], lg[1<<N];
int n, K;
int main(){
    //freopen("a.in", "r", stdin);
    //freopen("a.out", "w", stdout);
    cin>>n>>K;
    int x;
    FOR(i, 0, n){
        FOR(j, 0, (1 << i) - 1){
            scanf("%1d", &x);
            f[i][(1<<i)|j] = x;
            //printf("i=%d j=%d f=%d\n", i, j, f[i][(1<<i)|j]);
        }
    }
    FOR(i, 2, (1 << n) - 1)lg[i] = lg[i>>1] + 1;
    ROF(i, n, 0){
        FOR(s, 1 << i, (1 << i + 1) - 1){
            ROF(j, i, 0){
                if(!f[j][s])continue;
                int F = f[j][s];
                int s1 = s >> j;
                int s2 = s & ((1 << j) - 1);
                //printf("i=%d s=%d j=%d s1=%d s2=%d f=%d\n", i, s, j,  s1, s2, F);
                if(j)f[0][s1] += F;
                if(s2){
                    int k = lg[s2];
                    int S = ((s1 << 1 | 1) << k) | (s2 ^ (1 << k));
                    //printf("k=%d S=%d s1=%d\n", k, S, (s1 << 1 | 1) << k);
                    f[k][S] += F;
                }
                int s3 = (~s2) & ((1 << j) - 1);
                //printf("i=%d s=%d j=%d s1=%d s2=%d s3=%d F=%d\n", i, s, j, s1, s2, s3, F);
                if(s3){
                    int k = lg[s3];
                    int S = (s1 << k + 1) | (s2 & ((1 << k) - 1));
                    f[k][S] += F;
                }
            }
        }
    }
    ROF(i, n, 0){
        FOR(s, 0, (1 << i) - 1){
            if(f[0][s|(1<<i)] >= K){
                //printf("i=%d\n", i);
                for(int k = i - 1;k >= 0; k--){
                    printf("%d", s >> k & 1);
                }
                puts("");
                return 0;
            }
        }
    }
    return 0;
}