#CF2240D. 选择恐惧症 / D. Decidophobia

选择恐惧症 / D. Decidophobia

D. Decidophobia

Source: Codeforces 2240D
Contest: Codeforces Round 1105 (Div. 2)
Time limit: 2 seconds
Memory limit: 256 megabytes

Statement

There are nn people attending a round-table party, numbered 1,2,3,,n1, 2, 3, \ldots, n in clockwise order. You have prepared some gifts to distribute among them.

Each person ii has a weight aia_i and a common field of view dd. The field of view for person ii consists of the dd people sitting clockwise and the dd people sitting counter-clockwise from them (a total of 2d2d people excluding person ii).

The happiness gained by person ii is determined by the following rules:

  • If person ii receives a gift, and there are xx people within their field of view who did not receive a gift, they gain xaix \cdot a_i happiness.
  • If person ii does not receive a gift, and there are xx people within their field of view who received a gift, they incur xai-x \cdot a_i happiness.

You want to maximize the total happiness of all nn people combined. Find this maximum value.

Input

Each test contains multiple test cases. The first line contains the number of test cases tt (1t1041 \le t \le 10^4). The description of the test cases follows.

The first line of each test case contains two integers nn and dd (3n21053 \le n \le 2 \cdot 10^5, 1d<n21 \le d \lt \frac{n}{2}).

The second line contains nn integers a1,a2,,ana_1, a_2, \ldots, a_n (1ai1081 \le a_i \le 10^8), where aia_i is the weight of the ii-th person.

It is guaranteed that the sum of nn over all test cases does not exceed 10610^6.

Output

For each test case, output a single integer representing the maximum total happiness.

Note

In the first test case, there are 3 people sitting in a circle. For each person ii, the field of view is d=1d=1, which includes the 1 person sitting clockwise and the 1 person sitting counter-clockwise from them. If person 2 receives the gift, they gain happiness from 2 neighbors who did not receive it, resulting in 2a2=22=42 \cdot a_2 = 2 \cdot 2 = 4. However, person 1 and person 3 incur loss because they did not receive a gift but have a neighbor who did. After calculating all possibilities, the maximum total happiness is 3.

In the second test case, n=5,d=1n=5, d=1. The best solution is to give gifts to persons 2, 3, and 5, which can achieve the maximum total happiness value of 15.

Examples

5
3 1
1 2 3
5 1
1 4 5 2 6
6 2
1 1 4 5 1 4
10 2
230 24 3 42 432 234 934 2389 333 444
3 1
100000000 100000000 100000000
3
15
26
8590
0