#include <bits/stdc++.h>
using namespace std;
int read();
void print(int x) {
    if (x == 0) return printf("invalid string"), void();
    if (x == 1) return;
    print(x >> 1), putchar(x & 1 ^ '0');
}
int n, K;
char s[1100005];
int f[50000007], hi[2500005], lim = 1 << 21;
int nx[2500005][2];
void init() {
    for (int i = 2; i < lim; ++i) hi[i] = hi[i >> 1] + 1;
    for (int i = 2; i < lim; ++i) {
        nx[i >> 1][0] ? nx[i][0] = nx[i >> 1][0] << 1 | (i & 1) : 0;
        nx[i >> 1][1] ? nx[i][1] = nx[i >> 1][1] << 1 | (i & 1) : 0;
        !nx[i][i & 1] ? nx[i][i & 1] = 1 : 0;
    }
}
int id(int S, int T) { return (S << hi[T] | (T ^ (1 << hi[T]))) * 21 + hi[S]; }
int main() {
    n = read(), K = read(), init();
    for (int i = 0; i <= n; ++i) {
        scanf("%s", s);
        for (int j = 0, en = 1 << i; j < 1 << i; ++j)
            if (s[j] == '1') f[id(1, j | en)] = 1;
    }

    for (int T = (1 << n + 1) - 1; T > 1; --T) {
        int a = nx[T][0], b = nx[T][1];
        for (int S = (1 << n - hi[T] + 1) - 1; S; --S) {
            a ? f[id(S << 1, a)] += f[id(S, T)] : 0;
            b ? f[id(S << 1 | 1, b)] += f[id(S, T)] : 0;
            f[id(S, 1)] += f[id(S, T)];
        }
    }

    int res = 0;
    for (int S = (1 << n + 1) - 1; S; --S) {
        if (hi[S] < hi[res]) break;
        f[id(S, 1)] >= K ? res = S : 0;
    }
    print(res), puts("");
    return 0;
}
int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) f = (c == '-') ? -1 : f, c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
    return x * f;
}
### Prompt

Your challenge is to write a Cpp solution to 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;
int read();
void print(int x) {
    if (x == 0) return printf("invalid string"), void();
    if (x == 1) return;
    print(x >> 1), putchar(x & 1 ^ '0');
}
int n, K;
char s[1100005];
int f[50000007], hi[2500005], lim = 1 << 21;
int nx[2500005][2];
void init() {
    for (int i = 2; i < lim; ++i) hi[i] = hi[i >> 1] + 1;
    for (int i = 2; i < lim; ++i) {
        nx[i >> 1][0] ? nx[i][0] = nx[i >> 1][0] << 1 | (i & 1) : 0;
        nx[i >> 1][1] ? nx[i][1] = nx[i >> 1][1] << 1 | (i & 1) : 0;
        !nx[i][i & 1] ? nx[i][i & 1] = 1 : 0;
    }
}
int id(int S, int T) { return (S << hi[T] | (T ^ (1 << hi[T]))) * 21 + hi[S]; }
int main() {
    n = read(), K = read(), init();
    for (int i = 0; i <= n; ++i) {
        scanf("%s", s);
        for (int j = 0, en = 1 << i; j < 1 << i; ++j)
            if (s[j] == '1') f[id(1, j | en)] = 1;
    }

    for (int T = (1 << n + 1) - 1; T > 1; --T) {
        int a = nx[T][0], b = nx[T][1];
        for (int S = (1 << n - hi[T] + 1) - 1; S; --S) {
            a ? f[id(S << 1, a)] += f[id(S, T)] : 0;
            b ? f[id(S << 1 | 1, b)] += f[id(S, T)] : 0;
            f[id(S, 1)] += f[id(S, T)];
        }
    }

    int res = 0;
    for (int S = (1 << n + 1) - 1; S; --S) {
        if (hi[S] < hi[res]) break;
        f[id(S, 1)] >= K ? res = S : 0;
    }
    print(res), puts("");
    return 0;
}
int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) f = (c == '-') ? -1 : f, c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
    return x * f;
}
```