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

ll N, K;
ll a[1010];
vector<ll> cand;

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

int main() {
    cin >> N >> K;
    for (ll i = 0; i < N; i++) {
        cin >> a[i];
    }

    for (ll i = 0; i < N; i++) {
        ll sum = 0;
        for (ll j = i; j < N; j++) {
            sum += a[j];
            cand.push_back(sum);
        }
    }

    ll mask = 0;
    for (ll i = 40; i >= 0; i--) {
        if (countMask(mask | (1LL << i)) >= K) {
            mask |= (1LL << i);
        }
    }

    cout << mask << endl;

    return 0;
}

### 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;

ll N, K;
ll a[1010];
vector<ll> cand;

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

int main() {
    cin >> N >> K;
    for (ll i = 0; i < N; i++) {
        cin >> a[i];
    }

    for (ll i = 0; i < N; i++) {
        ll sum = 0;
        for (ll j = i; j < N; j++) {
            sum += a[j];
            cand.push_back(sum);
        }
    }

    ll mask = 0;
    for (ll i = 40; i >= 0; i--) {
        if (countMask(mask | (1LL << i)) >= K) {
            mask |= (1LL << i);
        }
    }

    cout << mask << endl;

    return 0;
}

```