#include <bits/stdc++.h>
 
using namespace std;
 
#define lli long long int
#define REP(i,s,l) for(lli i=s;i<l;i++)
#define DEBUG 0
#define INF (1LL<<50)
 
vector<lli> b;

lli countMask(lli a){
	lli ret=0;
	for(lli x:b) if((x&a)==a)ret++;
	return ret;
}

signed main(){
 
 	lli n,k;
 	cin>>n>>k;
 	vector<lli> a(n);
 	REP(i,0,n)cin>>a[i];

 	for(lli i=0;i<n;i++){
 		lli tmp = 0;
 		for(lli j=i;j<n;j++){
 			tmp += a[j];
 			b.push_back(tmp);
 		}
 	}

 	sort(b.begin(),b.end(),greater<lli>());

 	lli mask = 0;
 	for(lli i=40;i>=0;i--){
 		if(countMask(mask|(1LL<<i)) >= k){
 			mask |= (1LL<<i);
 		}
 	}
 	cout<<mask<<endl;

	return 0;
}
### Prompt

Please create a solution in Cpp to the following problem:

One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.

For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)

Find the maximum possible value for him.

Constraints

* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers

Input

Input is given from Standard Input in the following format:


N K
a_1 a_2 ... a_N


Output

Print the answer.

Examples

Input

4 2
2 5 2 5


Output

12


Input

8 4
9 1 8 2 7 5 6 4


Output

32

### Response

```cpp
#include <bits/stdc++.h>
 
using namespace std;
 
#define lli long long int
#define REP(i,s,l) for(lli i=s;i<l;i++)
#define DEBUG 0
#define INF (1LL<<50)
 
vector<lli> b;

lli countMask(lli a){
	lli ret=0;
	for(lli x:b) if((x&a)==a)ret++;
	return ret;
}

signed main(){
 
 	lli n,k;
 	cin>>n>>k;
 	vector<lli> a(n);
 	REP(i,0,n)cin>>a[i];

 	for(lli i=0;i<n;i++){
 		lli tmp = 0;
 		for(lli j=i;j<n;j++){
 			tmp += a[j];
 			b.push_back(tmp);
 		}
 	}

 	sort(b.begin(),b.end(),greater<lli>());

 	lli mask = 0;
 	for(lli i=40;i>=0;i--){
 		if(countMask(mask|(1LL<<i)) >= k){
 			mask |= (1LL<<i);
 		}
 	}
 	cout<<mask<<endl;

	return 0;
}
```