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

int N, K;
char X[22][1 << 20];
int dp[2][22][1 << 20];

int main() {
    scanf("%d %d", &N, &K);

    for(int i = 0; i <= N; i++) {
        scanf("\n");
        for(int j = 0; j < (1 << i); j++) {
            scanf("%c", &X[i][j]);
        }
    }

    for(int p = N; p >= 0; p--) {
        for(int len = 0; len <= N; len++) {
            for(int mask = 0; mask < (1 << max(p, len)); mask++) {
                dp[p & 1][len][mask] = 0;

                if(len <= p) {
                    dp[p & 1][len][mask] += X[p][mask] - '0';
                    if(p < N) {
                        dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len][mask];
                        dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len][mask ^ (1 << p)];
                    }
                    continue;
                }
                int t = (mask & (1 << p))? 1 : 0;
                int nmask = mask & ((1 << (N + 1)) - (1 << p));
                nmask <<= 1;
                nmask += (mask & ((1 << p) - 1));
                if(!t) nmask += (1 << p);
                dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len][mask];
                if(len < N) dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len + 1][nmask];
            }
        }
    }

    string ans = "";
    for(int i = N; i >= 0; i--) {
        for(int j = 0; j < (1 << i); j++) {
            if(dp[0][i][j] >= K) {
                string tmp;
                for(int k = i - 1; k >= 0; k--) {
                    tmp.push_back((j & (1 << k))? '1' : '0');
                }
                if(ans == "" || ans > tmp) ans = tmp;
            }
        }
        if(ans != "") {
            cout << ans;
            return 0;
        }
    }
}

### Prompt

Develop a solution in cpp 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<bits/stdc++.h>
using namespace std;

int N, K;
char X[22][1 << 20];
int dp[2][22][1 << 20];

int main() {
    scanf("%d %d", &N, &K);

    for(int i = 0; i <= N; i++) {
        scanf("\n");
        for(int j = 0; j < (1 << i); j++) {
            scanf("%c", &X[i][j]);
        }
    }

    for(int p = N; p >= 0; p--) {
        for(int len = 0; len <= N; len++) {
            for(int mask = 0; mask < (1 << max(p, len)); mask++) {
                dp[p & 1][len][mask] = 0;

                if(len <= p) {
                    dp[p & 1][len][mask] += X[p][mask] - '0';
                    if(p < N) {
                        dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len][mask];
                        dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len][mask ^ (1 << p)];
                    }
                    continue;
                }
                int t = (mask & (1 << p))? 1 : 0;
                int nmask = mask & ((1 << (N + 1)) - (1 << p));
                nmask <<= 1;
                nmask += (mask & ((1 << p) - 1));
                if(!t) nmask += (1 << p);
                dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len][mask];
                if(len < N) dp[p & 1][len][mask] += dp[ (p ^ 1) & 1 ][len + 1][nmask];
            }
        }
    }

    string ans = "";
    for(int i = N; i >= 0; i--) {
        for(int j = 0; j < (1 << i); j++) {
            if(dp[0][i][j] >= K) {
                string tmp;
                for(int k = i - 1; k >= 0; k--) {
                    tmp.push_back((j & (1 << k))? '1' : '0');
                }
                if(ans == "" || ans > tmp) ans = tmp;
            }
        }
        if(ans != "") {
            cout << ans;
            return 0;
        }
    }
}

```