#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 22;

vector<vector<int>> F[MAX_N];

vector<bool> exist[MAX_N];

int N, K;

void calc(int len, int val, int split) {
    int &res = F[len][val][split];
    if (split == 0) {
        res += exist[len][val];
        if (len < N) {
            res += F[len + 1][val][split];
            res += F[len + 1][val + (1 << len)][split];
        }
    } else {
        int pre = val & (1 << split) - 1, suf = val >> split;
        res += F[len][(pre >> 1) + (suf << split - 1) + ((pre & 1) << len - 1)][split - 1];
        if (len < N) {
            res += F[len + 1][val + (((pre & 1) ^ 1) << len)][split];
        }
    }
}

int main() {
    scanf("%d%d", &N, &K);
    for (int i = 0; i <= N; i++) {
        char buf[1 << MAX_N]; scanf("%s", buf);
        exist[i].resize(1 << i);
        for (int j = 0; j < 1 << i; j++) exist[i][j] = buf[j] == '1';
    }
    for (int i = 0; i <= N; i++) {
        F[i].resize(1 << i, vector<int>(i + 1));
    }
    for (int len = N; len >= 0; len--) {
        for (int split = 0; split <= len; split++) {
            for (int val = 0; val < 1 << len; val++) {
                calc(len, val, split);
            }
        }
    }
    for (int i = N; i >= 0; i--) {
        for (int j = 0; j < 1 << i; j++) {
            if (F[i][j][i] >= K) {
                for (int k = i - 1; k >= 0; k--)
                    printf("%d", j >> k & 1);
                printf("\n");
                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 <bits/stdc++.h>

using namespace std;

const int MAX_N = 22;

vector<vector<int>> F[MAX_N];

vector<bool> exist[MAX_N];

int N, K;

void calc(int len, int val, int split) {
    int &res = F[len][val][split];
    if (split == 0) {
        res += exist[len][val];
        if (len < N) {
            res += F[len + 1][val][split];
            res += F[len + 1][val + (1 << len)][split];
        }
    } else {
        int pre = val & (1 << split) - 1, suf = val >> split;
        res += F[len][(pre >> 1) + (suf << split - 1) + ((pre & 1) << len - 1)][split - 1];
        if (len < N) {
            res += F[len + 1][val + (((pre & 1) ^ 1) << len)][split];
        }
    }
}

int main() {
    scanf("%d%d", &N, &K);
    for (int i = 0; i <= N; i++) {
        char buf[1 << MAX_N]; scanf("%s", buf);
        exist[i].resize(1 << i);
        for (int j = 0; j < 1 << i; j++) exist[i][j] = buf[j] == '1';
    }
    for (int i = 0; i <= N; i++) {
        F[i].resize(1 << i, vector<int>(i + 1));
    }
    for (int len = N; len >= 0; len--) {
        for (int split = 0; split <= len; split++) {
            for (int val = 0; val < 1 << len; val++) {
                calc(len, val, split);
            }
        }
    }
    for (int i = N; i >= 0; i--) {
        for (int j = 0; j < 1 << i; j++) {
            if (F[i][j][i] >= K) {
                for (int k = i - 1; k >= 0; k--)
                    printf("%d", j >> k & 1);
                printf("\n");
                return 0;
            }
        }
    }
}

```