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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

#第5回 ドワンゴからの挑戦状 予選-B Sum AND Subarrays
"""
数列Aが与えられる。
整数kが与えられる。
数列Aの空でない部分列からk個取る。
各部分列の合計を"美しさ"と言う。
この時、k個すべての"美しさ"のbit andを取った時、最大値がいくつになるか
求めよ。
n <= 1000, k <= n*(n+1)//2

解法：
まず直感的に、すべての要素の合計sum(lst1)を使うとして良い。
また、回答は必ずこれ未満の値となる。
更に、andを用いることから、上位桁のbit積を優先して良い。

となると、結局大きい値が上位桁のbitも多いことから、
和が大きい順にk個取った時の論理積ということになる。

N<1000という制約から、n*(n+1)//2が愚直に間に合うので、これを
実装すれば良い。
"""
import sys
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0
n,k = map(int,readline().split())
lst1 = list(map(int,readline().split()))

def cumsum(lst): #元のリストを保持
    res = lst[:]
    for i in range(1,len(res)):
        res[i] += res[i-1]
    return res
def cumsumdif(l,r,lst): #求めたい差のインデックスを入れる
    if l == 0:
        return lst[r]
    else:
        return lst[r] - lst[l-1]
lst2 = cumsum(lst1)

lst3 = []

for i in range(n-1):
    for j in range(i,n):
        lst3.append(cumsumdif(i,j,lst2))

lst3.sort(reverse=True)

ans = lst3[0]
for i in range(1,k):
    ans = ans&lst3[i]

print(ans)