E. Kolya and Movie Theatre

Recently, Kolya found out that a new movie theatre is going to be opened in his city soon, which will show a new movie every day for $n$ days. So, on the day with the number $1 \le i \le n$, the movie theatre will show the premiere of the $i$-th movie. Also, Kolya found out the schedule of the movies and assigned the entertainment value to each movie, denoted by $a_i$.

However, the longer Kolya stays without visiting a movie theatre, the larger the decrease in entertainment value of the next movie. That decrease is equivalent to $d \cdot cnt$, where $d$ is a predetermined value and $cnt$ is the number of days since the last visit to the movie theatre. It is also known that Kolya managed to visit another movie theatre a day before the new one opened — the day with the number $0$. So if we visit the movie theatre the first time on the day with the number $i$, then $cnt$ — the number of days since the last visit to the movie theatre will be equal to $i$.

For example, if $d = 2$ and $a = [3, 2, 5, 4, 6]$, then by visiting movies with indices $1$ and $3$, $cnt$ value for the day $1$ will be equal to $1 - 0 = 1$ and $cnt$ value for the day $3$ will be $3 - 1 = 2$, so the total entertainment value of the movies will be $a_1 - d \cdot 1 + a_3 - d \cdot 2 = 3 - 2 \cdot 1 + 5 - 2 \cdot 2 = 2$.

Unfortunately, Kolya only has time to visit at most $m$ movies. Help him create a plan to visit the cinema in such a way that the total entertainment value of all the movies he visits is maximized.

Input

Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, and $d$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le n$, $1 \le d \le 10^9$).

The second line of each set of input data contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \le a_i \le 10^9$) — the entertainment values of the movies.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output

For each test case, output a single integer — the maximum total entertainment value that Kolya can get.

Example

input

6
5 2 2
3 2 5 4 6
4 3 2
1 1 1 1
6 6 6
-82 45 1 -77 39 11
5 2 2
3 2 5 4 8
2 1 1
-1 2
6 3 2
-8 8 -2 -1 9 0

output

Note

The first test case is explained in the problem statement.

In the second test case, it is optimal not to visit any movies.

In the third test case, it is optimal to visit movies with numbers $2$, $3$, $5$, $6$, so the total entertainment value of the visited movies will be $45 - 6 \cdot 2 + 1 - 6 \cdot 1 + 39 - 6 \cdot 2 + 11 - 6 \cdot 1 = 60$.

解题思路

　　没发现式子的性质没做出来，好似喵。

　　假设从小到大选择了$k$个下标$i_1, i_2, \ldots, i_k$，此时的结果是$$\begin{align*}&(a_{i_1} - d \cdot i_1) + (a_{i_2} - d \cdot (i_2 - i_1)) + \ldots + (a_{i_k} - d \cdot (i_k - i_{k-1})) \\ = \ &(a_{i_1} + a_{i_2} + \ldots + a_{i_k}) - d \cdot (i_1 + i_2 - i_1 + i_3 - i_2 + \ldots + i_k - i_{k-1}) \\ = \ &(a_{i_1} + a_{i_2} + \ldots + a_{i_k}) - d \cdot i_k \end{align*}$$

　　可以发现结果中减去的部分只取决于最大的下标，因此可以从小到大枚举最后一个下标$k$选什么，然后再从前面的下标$i \in [1, k-1]$中选出最大的$m-1$个$a_i$（此时已经选择了$a_k$），来使得结果最大。因此可以开个堆来维护前缀的前$m-1$个最大值以及他们的和。需要注意的是，如果$a_k \leq 0$那么我们永远不会选择这个元素，从上面的式子可以看出，如果$k$不作为最大的下标，那么我们将$a_k$删除结果不会变小，而如果$k$作为最大的下标，那么删除$a_k$且$d \cdot i_k$变小，结果会变小。总之删除$a_k$结果一定不会变小。

　　AC代码如下，时间复杂度为$O(n \log {m})$：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 
 6 const int N = 2e5 + 10;
 7 
 8 int a[N];
 9 
10 void solve() {
11     int n, m, d;
12     scanf("%d %d %d", &n, &m, &d);
13     for (int i = 1; i <= n; i++) {
14         scanf("%d", a + i);
15     }
16     priority_queue<int, vector<int>, greater<int>> pq;
17     LL ret = 0, s = 0;
18     for (int i = 1; i <= n; i++) {
19         if (a[i] <= 0) continue;
20         s += a[i];
21         if (pq.size() == m) {
22             s -= pq.top();
23             pq.pop();
24         }
25         pq.push(a[i]);
26         ret = max(ret, s - 1ll * i * d);
27     }
28     printf("%lld\n", ret);
29 }
30 
31 int main() {
32     int t;
33     scanf("%d", &t);
34     while (t--) {
35         solve();
36     }
37     
38     return 0;
39 }