#include<bits/stdc++.h>
using namespace std;
const int maxn=1<<22;
typedef long long ll;
typedef unsigned long long ull;
int data[maxn][21],n,k;
int mem[maxn*2];
char s[maxn];
int mx=1000000000,mxl=0;
void update(int sta,int nums,int len) {
    mem[sta|(1<<len)]+=nums;
    if(mem[sta|(1<<len)]>=k)
        if(mxl<len)mxl=len,mx=sta;
        else if(mxl==len)mx=min(mx,sta);

}
int main() {
    scanf("%d%d",&n,&k);
    for(int i=0;i<=n;i++) {
        scanf("%s",s);//len=2^i
        for(int j=0;j<(1<<i);j++) {
            int tmp=j;
            tmp|=1<<i;
            data[tmp][0]+=(s[j]=='1');
        }
    }
    for(int i=0;i<=n;i++) {
        for(int j=0;j<(1<<(n+1));j++) {
            if(!data[j][i])continue;//t-s
            ull s=j&((1<<i)-1),t=j>>i;int l=(32-__builtin_clz(t))-1;//有t多少位
            if(i)update(s,data[j][i],i);
            if(!l){continue;}
            t^=(1<<l);
            int first1=(32-(t?__builtin_clz(t):32));
            unsigned int num=t<<(32-l);
            int first0=l?(l-((~num)?__builtin_clz(~num):l)):0;
            int newt,news;
            if(first1>0){
                newt=t&((1<<first1)-1);
                newt|=(1<<first1-1);
                news=(s<<1)|1;
                data[(newt<<(i+1))|news][i+1]+=data[j][i];
            }
            if(first0>0){
                newt=t&((1<<first0)-1);
                newt|=(1<<first0-1);
                news=(s<<1);
                data[(newt<<(i+1))|news][i+1]+=data[j][i];
            }
        }
    }
    if(mx==1000000000){puts("");return 0;}
    for(int i=mxl;i>=1;i--) {
        printf("%d",(mx&(1<<i-1))?1:0);
    }
    printf("\n");
    return 0;
}
### Prompt

Your task is to create 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;
const int maxn=1<<22;
typedef long long ll;
typedef unsigned long long ull;
int data[maxn][21],n,k;
int mem[maxn*2];
char s[maxn];
int mx=1000000000,mxl=0;
void update(int sta,int nums,int len) {
    mem[sta|(1<<len)]+=nums;
    if(mem[sta|(1<<len)]>=k)
        if(mxl<len)mxl=len,mx=sta;
        else if(mxl==len)mx=min(mx,sta);

}
int main() {
    scanf("%d%d",&n,&k);
    for(int i=0;i<=n;i++) {
        scanf("%s",s);//len=2^i
        for(int j=0;j<(1<<i);j++) {
            int tmp=j;
            tmp|=1<<i;
            data[tmp][0]+=(s[j]=='1');
        }
    }
    for(int i=0;i<=n;i++) {
        for(int j=0;j<(1<<(n+1));j++) {
            if(!data[j][i])continue;//t-s
            ull s=j&((1<<i)-1),t=j>>i;int l=(32-__builtin_clz(t))-1;//有t多少位
            if(i)update(s,data[j][i],i);
            if(!l){continue;}
            t^=(1<<l);
            int first1=(32-(t?__builtin_clz(t):32));
            unsigned int num=t<<(32-l);
            int first0=l?(l-((~num)?__builtin_clz(~num):l)):0;
            int newt,news;
            if(first1>0){
                newt=t&((1<<first1)-1);
                newt|=(1<<first1-1);
                news=(s<<1)|1;
                data[(newt<<(i+1))|news][i+1]+=data[j][i];
            }
            if(first0>0){
                newt=t&((1<<first0)-1);
                newt|=(1<<first0-1);
                news=(s<<1);
                data[(newt<<(i+1))|news][i+1]+=data[j][i];
            }
        }
    }
    if(mx==1000000000){puts("");return 0;}
    for(int i=mxl;i>=1;i--) {
        printf("%d",(mx&(1<<i-1))?1:0);
    }
    printf("\n");
    return 0;
}
```