#include <cstdio>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;

int n,k,num=0;
ll a[1005],s[1005],ans=0;
vector <ll> d;

int main(void){
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++)
	scanf("%lld",&a[i]),s[i]=s[i-1]+a[i];
	for(int i=1;i<=n;i++){
		for(int j=i;j<=n;j++)
		d.push_back(s[j]-s[i-1]);
	}
	for(int i=50;i>=0;i--){
		vector <ll> e;
		int cnt=0;
		for(int j=0;j<d.size();j++){
			ll x=d[j];
			if((x>>i)&1){
				e.push_back(x);
				cnt++;
			}	
		}
		if(cnt>=k){
			ans+=(1ll<<i);
			d=e;
		}
	}
	printf("%lld",ans);
}
/*
1-->1
2-->10
3-->11
5-->101
*/ 
### Prompt

In CPP, your task is to solve 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 <cstdio>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;

int n,k,num=0;
ll a[1005],s[1005],ans=0;
vector <ll> d;

int main(void){
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++)
	scanf("%lld",&a[i]),s[i]=s[i-1]+a[i];
	for(int i=1;i<=n;i++){
		for(int j=i;j<=n;j++)
		d.push_back(s[j]-s[i-1]);
	}
	for(int i=50;i>=0;i--){
		vector <ll> e;
		int cnt=0;
		for(int j=0;j<d.size();j++){
			ll x=d[j];
			if((x>>i)&1){
				e.push_back(x);
				cnt++;
			}	
		}
		if(cnt>=k){
			ans+=(1ll<<i);
			d=e;
		}
	}
	printf("%lld",ans);
}
/*
1-->1
2-->10
3-->11
5-->101
*/ 
```