#include <bits/stdc++.h>
using namespace std;
#define N 21
int f[N][1<<N],nex[1<<N][2],len[1<<N];
inline int read() {
    int x=0;
    char ch=getchar();
    while (!isdigit(ch)) ch=getchar();
    while (isdigit(ch)) x=x*10+ch-'0',ch=getchar();
    return x;
}
inline char Getchar() {
    char ch=getchar();
    while (ch!='1' && ch!='0') ch=getchar();
    return ch;
}
int main() {
    int n=read(),k=read(),lim=2<<n;
    for (int i=0;i<=n;i++) 
        for (int j=0;j<1<<i;j++) f[i][1<<i^j]=Getchar()-'0';
    len[0]=-1;
    for (int i=1;i<lim;i++) len[i]=len[i>>1]+1;
    for (int i=2;i<lim;i++) {
        if (i>>(len[i]-1)&1) nex[i][1]=i^1<<len[i],nex[i][0]=nex[nex[i][1]][0]/* ,printf("%d %d\n",i,nex[i][1]) */;
        else nex[i][0]=i^3<<(len[i]-1),nex[i][1]=nex[nex[i][0]][1];
        // printf("%d %d %d\n",i,nex[i][0],nex[i][1]);
    }
    // for (int i=2;i<lim;i++) 
    for (int i=n;i>=1;i--)
        for (int A=1<<i;A<lim;A++) {
            if (!f[i][A]) continue;
            int C=A>>i,B=(A&((1<<i)-1))^1<<i;
            if (nex[B][0]) f[len[nex[B][0]]][(C<<1^1)<<len[nex[B][0]]^nex[B][0]]+=f[i][A];
            if (nex[B][1]) f[len[nex[B][1]]][(C<<1^0)<<len[nex[B][1]]^nex[B][1]]+=f[i][A];
            f[0][C]+=f[i][A];
            // printf("%d %d %d\n",i,A,B);
        }
    int ans=1;
    for (int i=2;i<lim;i++)
        if (f[0][i]>=k && len[i]>len[ans]) ans=i;
    // printf("%d\n",ans);
	while (~ans>>n&1) --n;
	while (n--) printf("%d",ans>>n&1);
	puts("");
    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;
#define N 21
int f[N][1<<N],nex[1<<N][2],len[1<<N];
inline int read() {
    int x=0;
    char ch=getchar();
    while (!isdigit(ch)) ch=getchar();
    while (isdigit(ch)) x=x*10+ch-'0',ch=getchar();
    return x;
}
inline char Getchar() {
    char ch=getchar();
    while (ch!='1' && ch!='0') ch=getchar();
    return ch;
}
int main() {
    int n=read(),k=read(),lim=2<<n;
    for (int i=0;i<=n;i++) 
        for (int j=0;j<1<<i;j++) f[i][1<<i^j]=Getchar()-'0';
    len[0]=-1;
    for (int i=1;i<lim;i++) len[i]=len[i>>1]+1;
    for (int i=2;i<lim;i++) {
        if (i>>(len[i]-1)&1) nex[i][1]=i^1<<len[i],nex[i][0]=nex[nex[i][1]][0]/* ,printf("%d %d\n",i,nex[i][1]) */;
        else nex[i][0]=i^3<<(len[i]-1),nex[i][1]=nex[nex[i][0]][1];
        // printf("%d %d %d\n",i,nex[i][0],nex[i][1]);
    }
    // for (int i=2;i<lim;i++) 
    for (int i=n;i>=1;i--)
        for (int A=1<<i;A<lim;A++) {
            if (!f[i][A]) continue;
            int C=A>>i,B=(A&((1<<i)-1))^1<<i;
            if (nex[B][0]) f[len[nex[B][0]]][(C<<1^1)<<len[nex[B][0]]^nex[B][0]]+=f[i][A];
            if (nex[B][1]) f[len[nex[B][1]]][(C<<1^0)<<len[nex[B][1]]^nex[B][1]]+=f[i][A];
            f[0][C]+=f[i][A];
            // printf("%d %d %d\n",i,A,B);
        }
    int ans=1;
    for (int i=2;i<lim;i++)
        if (f[0][i]>=k && len[i]>len[ans]) ans=i;
    // printf("%d\n",ans);
	while (~ans>>n&1) --n;
	while (n--) printf("%d",ans>>n&1);
	puts("");
    return 0;
}

```