#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

int main(){
	
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n, k;
	cin >> n >> k;
	vector <int> v(n);
	for(int i = 0; i < n; i++){
		cin >> v[i];
	}	
	vector <ll> vec;
	for(int i = 0; i < n; i++){
		ll curr = 0;
		for(int j = i; j < n; j++){
			curr += v[j];
			vec.push_back(curr);
		}
	}
	ll res = 0;
	for(int i = 63; i >= 0; i--){
		int cnt = 0;
		for(auto x : vec){
			if(x & (1LL << i)) cnt++;
		}
		if(cnt >= k){
			res += (1LL << i);
			vector <ll> next;
			for(auto x : vec){
				if(!(x & (1LL << i))) continue;
				next.push_back(x);
			}
			vec = next;
		}
	}
	cout << res;
}
### 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;

typedef long long ll;

int main(){
	
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n, k;
	cin >> n >> k;
	vector <int> v(n);
	for(int i = 0; i < n; i++){
		cin >> v[i];
	}	
	vector <ll> vec;
	for(int i = 0; i < n; i++){
		ll curr = 0;
		for(int j = i; j < n; j++){
			curr += v[j];
			vec.push_back(curr);
		}
	}
	ll res = 0;
	for(int i = 63; i >= 0; i--){
		int cnt = 0;
		for(auto x : vec){
			if(x & (1LL << i)) cnt++;
		}
		if(cnt >= k){
			res += (1LL << i);
			vector <ll> next;
			for(auto x : vec){
				if(!(x & (1LL << i))) continue;
				next.push_back(x);
			}
			vec = next;
		}
	}
	cout << res;
}
```