#include <cstdio>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#define fi first
#define se second
using namespace std;
typedef long long i64;
typedef pair<int, i64> pii;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 21;
int n, m, f[N][1<<N];
int code(int i, int k, int u, int S, int T, int p)
{
    return (1<<(i-k+u+1))|(S<<(u+1))|(T&((1<<u)-1))|(p<<u);
}
int main()
{
    n = read(), m = read();
    for(int i = 0; i <= n; ++i) for(int j = 0; j < 1<<i; ++j) scanf("%1d", &f[i][(1<<i)|j]);
    for(int i = n; ~i; --i)
        for(int j = 0; j < 1<<i; ++j)
            for(int k = i; ~k; --k)
            {
                int w = f[k][(1<<i)|j]; if(!w) continue;
                int S = j>>k, T = j&((1<<k)-1), u;
                if(k) f[0][(1<<i-k)|S] += w;
                if(T)
                {
                    for(u = k-1; ~T>>u&1; --u);
                    f[u][code(i, k, u, S, T, 1)] += w;
                }
                if(~T&((1<<k)-1))
                {
                    for(u = k-1; T>>u&1; --u);
                    f[u][code(i, k, u, S, T, 0)] += w;                    
                }
            }
    for(int i = n; ~i; --i)
        for(int j = 0; j < 1<<i; ++j)
            if(f[0][(1<<i)|j] >= m)
            {
                for(int k = i-1; ~k; --k) putchar('0'+(j>>k&1));
                return puts(""), 0;
            }
    return 0;
}
### Prompt

In CPP, your task is to solve the following 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

### Response

```cpp
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#define fi first
#define se second
using namespace std;
typedef long long i64;
typedef pair<int, i64> pii;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 21;
int n, m, f[N][1<<N];
int code(int i, int k, int u, int S, int T, int p)
{
    return (1<<(i-k+u+1))|(S<<(u+1))|(T&((1<<u)-1))|(p<<u);
}
int main()
{
    n = read(), m = read();
    for(int i = 0; i <= n; ++i) for(int j = 0; j < 1<<i; ++j) scanf("%1d", &f[i][(1<<i)|j]);
    for(int i = n; ~i; --i)
        for(int j = 0; j < 1<<i; ++j)
            for(int k = i; ~k; --k)
            {
                int w = f[k][(1<<i)|j]; if(!w) continue;
                int S = j>>k, T = j&((1<<k)-1), u;
                if(k) f[0][(1<<i-k)|S] += w;
                if(T)
                {
                    for(u = k-1; ~T>>u&1; --u);
                    f[u][code(i, k, u, S, T, 1)] += w;
                }
                if(~T&((1<<k)-1))
                {
                    for(u = k-1; T>>u&1; --u);
                    f[u][code(i, k, u, S, T, 0)] += w;                    
                }
            }
    for(int i = n; ~i; --i)
        for(int j = 0; j < 1<<i; ++j)
            if(f[0][(1<<i)|j] >= m)
            {
                for(int k = i-1; ~k; --k) putchar('0'+(j>>k&1));
                return puts(""), 0;
            }
    return 0;
}
```