#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <set>
#include <map>
#include <cassert>
#include <numeric>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long int int64;

const int LOG = 20;
const int N = (1 << LOG) + 10;
char str[N];

int ar[LOG + 1];
int arsz;

int dp[LOG + 1][1 << (LOG + 1)];

int main(int,  char **)
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 0; i <= n; i++)
    {
        scanf("%s", str);
        for (int j = 0; j < (1 << i); j++)
        {
            if (str[j] == '0') continue;
            dp[0][j | (1 << i)]++;
        }
    }
    for (int t = 0; t < LOG; t++)
    {
        eprintf("--- t = %d\n", t);
        int pos = 0;
        for (int i = 0; i < (1 << (LOG + 1)); i++)
        {
            int cur = dp[t][i];
            if (cur == 0) continue;
            while ((1 << (pos + 1)) <= i) pos++;
            int bnd = pos - t - 1;
            if (bnd < 0) continue;
            eprintf("i = %d, val = %d\n", i, cur);

            int x = (1 << (bnd + 1));
            int gen = (i >> (bnd + 1));
            int msk = (i & (x - 1));
            int b = (msk >> bnd) & 1;
            while (bnd >= 0 && ((msk >> bnd) & 1) == b)
                bnd--;
            
            dp[t][gen] += cur;
            dp[t + 1][i] += cur;
            eprintf("-> %d ", i);
            if (bnd >= 0)
            {
                int oth = ( ( (gen << 1) ^ 1 ^ b) << bnd) | (msk & ((1 << bnd) - 1));
                dp[t + 1][oth] += cur;
                eprintf("%d", oth);
            }
            eprintf("\n");
        }
    }
    for (int t = LOG; t >= 0; t--)
        for (int i = 0; i < (1 << (LOG + 1)); i++)
            if (dp[t][i] >= k)
            {
                for (int j = t - 1; j >= 0; j--)
                    printf("%d", (i >> j) & 1);
                printf("\n");
                return 0;
            }




	return 0;
}


### Prompt

Please provide a cpp coded solution to the problem described below:

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 <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <set>
#include <map>
#include <cassert>
#include <numeric>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long int int64;

const int LOG = 20;
const int N = (1 << LOG) + 10;
char str[N];

int ar[LOG + 1];
int arsz;

int dp[LOG + 1][1 << (LOG + 1)];

int main(int,  char **)
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 0; i <= n; i++)
    {
        scanf("%s", str);
        for (int j = 0; j < (1 << i); j++)
        {
            if (str[j] == '0') continue;
            dp[0][j | (1 << i)]++;
        }
    }
    for (int t = 0; t < LOG; t++)
    {
        eprintf("--- t = %d\n", t);
        int pos = 0;
        for (int i = 0; i < (1 << (LOG + 1)); i++)
        {
            int cur = dp[t][i];
            if (cur == 0) continue;
            while ((1 << (pos + 1)) <= i) pos++;
            int bnd = pos - t - 1;
            if (bnd < 0) continue;
            eprintf("i = %d, val = %d\n", i, cur);

            int x = (1 << (bnd + 1));
            int gen = (i >> (bnd + 1));
            int msk = (i & (x - 1));
            int b = (msk >> bnd) & 1;
            while (bnd >= 0 && ((msk >> bnd) & 1) == b)
                bnd--;
            
            dp[t][gen] += cur;
            dp[t + 1][i] += cur;
            eprintf("-> %d ", i);
            if (bnd >= 0)
            {
                int oth = ( ( (gen << 1) ^ 1 ^ b) << bnd) | (msk & ((1 << bnd) - 1));
                dp[t + 1][oth] += cur;
                eprintf("%d", oth);
            }
            eprintf("\n");
        }
    }
    for (int t = LOG; t >= 0; t--)
        for (int i = 0; i < (1 << (LOG + 1)); i++)
            if (dp[t][i] >= k)
            {
                for (int j = t - 1; j >= 0; j--)
                    printf("%d", (i >> j) & 1);
                printf("\n");
                return 0;
            }




	return 0;
}


```