#include <bits/stdc++.h>
using namespace std ;
#define ll long long 
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 1000, NN = N * N;
int fr[60], n, k, y[N];
vector<ll> x;
bool add[NN];

void ad(ll x, int v){
	int j = 0;
	while (x){
		if (x & 1)fr[j] += v;
		++j;
		x >>= 1;
	}
}

ll go(){
	ll an = 0;
	for (int i = 59; i >= 0; --i)if (fr[i] >= k){
		an |= 1ll << i;
		f(j, 0, x.size())if (add[j] && !(x[j] >> i & 1))ad(x[j], -1), add[j] = false;
	}
	return an;
}

int main(){
	scanf("%d%d", &n, &k);
	f(i, 0, n)scanf("%d", y + i);
	f(i, 0, n){
		ll s = 0;
		f(j, i, n)s += y[j], x.push_back(s);
	}
	f(i, 0, x.size())add[i] = true;
	f(i, 0, x.size())ad(x[i], 1);
	printf("%lld\n", go());
}
### Prompt

Please provide a CPP coded solution 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 ll long long 
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 1000, NN = N * N;
int fr[60], n, k, y[N];
vector<ll> x;
bool add[NN];

void ad(ll x, int v){
	int j = 0;
	while (x){
		if (x & 1)fr[j] += v;
		++j;
		x >>= 1;
	}
}

ll go(){
	ll an = 0;
	for (int i = 59; i >= 0; --i)if (fr[i] >= k){
		an |= 1ll << i;
		f(j, 0, x.size())if (add[j] && !(x[j] >> i & 1))ad(x[j], -1), add[j] = false;
	}
	return an;
}

int main(){
	scanf("%d%d", &n, &k);
	f(i, 0, n)scanf("%d", y + i);
	f(i, 0, n){
		ll s = 0;
		f(j, i, n)s += y[j], x.push_back(s);
	}
	f(i, 0, x.size())add[i] = true;
	f(i, 0, x.size())ad(x[i], 1);
	printf("%lld\n", go());
}
```