#include <bits/stdc++.h>
using namespace std;

const int N = 21;

// 最高位代表其长度 1 << i -> 长度为 i 

int *dp[1 << N], tp[2][1 << N], fir[1 << N]; char s[1 << N];

int main() {
    int n, k; scanf("%d%d", &n, &k);
    dp[1] = new int[1 << n + 1], dp[1][0] = 0;
    for (int i = 1; i < 1 << n + 1; i++) fir[i] = fir[i >> 1] + 1;
    for (int i = 0; i <= n; i++) {
        scanf("%s", s); int base = 1 << i;
        for (int j = 0; j < base; j++) 
            dp[1][base | j] = s[j] - '0';
        int all = base - 1;
        for (int j = 0; j < base; j++) 
            tp[1][base | j] = j;
        tp[0][base * 2 - 1] = 0;
        for (int j = 0; j + 1 < base; j++) {
            int t = all ^ j, pos = fir[t]; 
            tp[0][base | j] = (1 << pos - 1) | (j & ((1 << pos - 1) - 1));
        }
    }
    int res = 0;
    for (int i = 1; i <= n; i++) {
        int base = 1 << i;
        for (int s = base - 1; s >= 0; s--) {
            dp[base + s] = new int[1 << n + 1 - i];
            for (int j = 0; j < (1 << n + 1 - i); j++) dp[base + s][j] = 0;
            for (int j = 1; j < 1 << (n + 1 - (i - 1)); j++) if (tp[s & 1][j])
                dp[base + s][tp[s & 1][j]] += dp[(base + s) >> 1][j];
            int tot = 0;
            for (int j = 1; j < (1 << n + 1 - i); j++) tot += dp[base + s][j];
            if (tot >= k) res = base + s;
        }
    }
    int len = fir[res];
    for (int i = len - 1; i >= 1; i--) {
        if (res & (1 << i - 1)) putchar('1');
        else putchar('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 <bits/stdc++.h>
using namespace std;

const int N = 21;

// 最高位代表其长度 1 << i -> 长度为 i 

int *dp[1 << N], tp[2][1 << N], fir[1 << N]; char s[1 << N];

int main() {
    int n, k; scanf("%d%d", &n, &k);
    dp[1] = new int[1 << n + 1], dp[1][0] = 0;
    for (int i = 1; i < 1 << n + 1; i++) fir[i] = fir[i >> 1] + 1;
    for (int i = 0; i <= n; i++) {
        scanf("%s", s); int base = 1 << i;
        for (int j = 0; j < base; j++) 
            dp[1][base | j] = s[j] - '0';
        int all = base - 1;
        for (int j = 0; j < base; j++) 
            tp[1][base | j] = j;
        tp[0][base * 2 - 1] = 0;
        for (int j = 0; j + 1 < base; j++) {
            int t = all ^ j, pos = fir[t]; 
            tp[0][base | j] = (1 << pos - 1) | (j & ((1 << pos - 1) - 1));
        }
    }
    int res = 0;
    for (int i = 1; i <= n; i++) {
        int base = 1 << i;
        for (int s = base - 1; s >= 0; s--) {
            dp[base + s] = new int[1 << n + 1 - i];
            for (int j = 0; j < (1 << n + 1 - i); j++) dp[base + s][j] = 0;
            for (int j = 1; j < 1 << (n + 1 - (i - 1)); j++) if (tp[s & 1][j])
                dp[base + s][tp[s & 1][j]] += dp[(base + s) >> 1][j];
            int tot = 0;
            for (int j = 1; j < (1 << n + 1 - i); j++) tot += dp[base + s][j];
            if (tot >= k) res = base + s;
        }
    }
    int len = fir[res];
    for (int i = len - 1; i >= 1; i--) {
        if (res & (1 << i - 1)) putchar('1');
        else putchar('0');
    }
    return 0;
}

```