Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 2
2 5 2 5

Sample Output 1

12

There are 10 nonempty contiguous subsequences of a. Let us enumerate them:

contiguous subsequences starting from the first element: \{2\}, \{2, 5\}, \{2, 5, 2\}, \{2, 5, 2, 5\}

contiguous subsequences starting from the second element: \{5\}, \{5, 2\}, \{5, 2, 5\}

contiguous subsequences starting from the third element: \{2\}, \{2, 5\}

contiguous subsequences starting from the fourth element: \{5\}

(Note that even if the elements of subsequences are equal, subsequences that have different starting indices are considered to be different.)

The maximum possible bitwise AND of the beauties of two different contiguous subsequences is 12.
This can be achieved by choosing \{5, 2, 5\} (with beauty 12) and \{2, 5, 2, 5\} (with beauty 14).

Sample Input 2

8 4
9 1 8 2 7 5 6 4

Sample Output 2

32

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
const long long INF = 1LL << 60;
const long long MOD = 1000000007;
const double PI = acos(-1.0);
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rep1(i, n) for (ll i = 1; i <= (n); ++i)
#define rrep(i, n) for (ll i = (n - 1); i >= 0; --i)
#define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c)))
#define ALL(obj) (obj).begin(), (obj).end()
#define RALL(obj) (obj).rbegin(), (obj).rend()
#define pb push_back
#define to_s to_string
#define len(v) (ll)v.size()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define print(x) cout << (x) << '\n'
#define drop(x) cout << (x) << '\n', exit(0)
#define debug(x) cout << #x << ": " << (x) << '\n'
using namespace std;
using ll = long long;
typedef pair<ll, ll> P;
typedef vector<ll> vec;
typedef vector<vector<ll>> vec2;
typedef vector<vector<vector<ll>>> vec3;
template<class S, class T> inline bool chmax(S &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class S, class T> inline bool chmin(S &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline ll msb(ll v) { return 1 << (31 - __builtin_clzll(v)); }
inline ll devc(ll x, ll y) { return (x + y - 1) / y; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }

struct IoSetup {
    IoSetup() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
} iosetup;

template< typename T1, typename T2 >
ostream &operator << (ostream &os, const pair< T1, T2 > &p) {
    os << p.first << " " << p.second;
    return os;
}

template< typename T1, typename T2 >
istream &operator >> (istream &is, pair< T1, T2 > &p) {
    is >> p.first >> p.second;
    return is;
}

template< typename T1, typename T2, typename T3 >
ostream &operator << (ostream &os, const tuple< T1, T2, T3 > &t) {
    os << get<0>(t) << " " << get<1>(t) << " " << get<2>(t);
    return os;
}

template< typename T1, typename T2, typename T3 >
istream &operator >> (istream &is, tuple< T1, T2, T3 > &t) {
    is >> get<0>(t) >> get<1>(t) >> get<2>(t);
    return is;
}

template< typename T >
ostream &operator << (ostream &os, const vector< T > &v){
    for (int i = 0; i < (int)v.size(); ++i) {
        os << v[i] << (i + 1 != v.size() ? " " : "");
    }
    return os;
}

template< typename T >
istream &operator >>  (istream &is, vector< T > &v){
    for(T &in : v) is >> in;
    return is;
}

/*---------------------------------    Tools    ------------------------------------------*/
template< typename T >
vector<T> cumsum(const vector<T> &X){
    vector<T> res(X.size() + 1, 0);
    for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i];
    return res;
}

template< typename S, typename T, typename F>
pair<T, T> bisearch(S left, T right, F f) {
    while(abs(right - left) > 1){
        T mid = (right + left) / 2;
        if(f(mid)) right = mid;
        else left = mid;
    }
    return {left, right};
}

template< typename S, typename T, typename F>
double trisearch(S left, T right, F f, int maxLoop = 90){
    double low = left, high = right;
    while(maxLoop--){
        double mid_left = high / 3 + low * 2 / 3;
        double mid_right = high * 2 / 3 + low / 3;
        if(f(mid_left) >= f(mid_right)) low = mid_left;
        else high = mid_right;
    }
    return (low + high) * 0.5;
}

/*------------------------------- Main Code Here -----------------------------------------*/
/* https://atcoder.jp/contests/dwacon5th-prelims/tasks/dwacon5th_prelims_b */

int main()
{
    ll N, K;
    cin >> N >> K;
    vec A(N);
    cin >> A;

    vec cum = cumsum(A);

    map<string, ll, greater<>> mp;

    rep1(i, N) rep1(j, N) if(i <= j){
        bitset<31> beauty = cum[j] - cum[i - 1];
        mp[beauty.to_string()]++;
    }

    bitset<31> ans = ~0;
    for(auto m : mp){
        if(K <= 0) break;
        bitset<31> now(m.first);
        ll ct = m.second;
        ans &= now;
        K -= ct;
    }

    print(ans.to_ullong());

    return 0;
}