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

#define F first
#define S second
#define PB push_back
#define all(x) x.begin(), x.end()

typedef long long ll; 
typedef double ld;

const int N = 1000 + 10, M = 1000 + 10; 

int a[N];

ll solve(vector<ll>& v, ll id, ll k)
{
	if(id==-1) return 0;
	vector<ll> v2;
	for(auto u : v)
		if(u&(1ll<<id))
			v2.PB(u);
	if(v2.size()>=k) return (1ll<<id) + solve(v2,id-1,k);
	return solve(v,id-1,k);
}

int32_t main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n, k;
	cin>> n >> k;
	vector<ll> v;
	for(int i = 0; i < n; i++)
	{
		cin>> a[i];
		ll sum = 0;
		for(int j = i; j > -1; j--) sum += a[j], v.PB(sum);
	}
		
	cout<< solve(v, 50, k) << endl;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 F first
#define S second
#define PB push_back
#define all(x) x.begin(), x.end()

typedef long long ll; 
typedef double ld;

const int N = 1000 + 10, M = 1000 + 10; 

int a[N];

ll solve(vector<ll>& v, ll id, ll k)
{
	if(id==-1) return 0;
	vector<ll> v2;
	for(auto u : v)
		if(u&(1ll<<id))
			v2.PB(u);
	if(v2.size()>=k) return (1ll<<id) + solve(v2,id-1,k);
	return solve(v,id-1,k);
}

int32_t main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n, k;
	cin>> n >> k;
	vector<ll> v;
	for(int i = 0; i < n; i++)
	{
		cin>> a[i];
		ll sum = 0;
		for(int j = i; j > -1; j--) sum += a[j], v.PB(sum);
	}
		
	cout<< solve(v, 50, k) << endl;
}
```