CF1684D.Traps

There are $n$ traps numbered from $1$ to $n$ . You will go through them one by one in order. The $i$ -th trap deals $a_i$ base damage to you.

Instead of going through a trap, you can jump it over. You can jump over no more than $k$ traps. If you jump over a trap, it does not deal any damage to you. But there is an additional rule: if you jump over a trap, all next traps damages increase by $1$ (this is a bonus damage).

Note that if you jump over a trap, you don't get any damage (neither base damage nor bonus damage). Also, the bonus damage stacks so, for example, if you go through a trap $i$ with base damage $a_i$ , and you have already jumped over $3$ traps, you get $(a_i + 3)$ damage.

You have to find the minimal damage that it is possible to get if you are allowed to jump over no more than $k$ traps.

The input consists of multiple test cases. The first line contains a single integer $t$ ( $1 \le t \le 100$ ) — the number of test cases. Description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ( $1 \le n \le 2 \cdot 10^5$ , $1 \le k \le n$ ) — the number of traps and the number of jump overs that you are allowed to make.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \le a_i \le 10^9$ ) — base damage values of all traps.

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

For each test case output a single integer — the minimal total damage that it is possible to get if you are allowed to jump over no more than $k$ traps.

In the first test case it is allowed to jump over all traps and take $0$ damage.

In the second test case there are $5$ ways to jump over some traps:

1. Do not jump over any trap.Total damage: $5 + 10 + 11 + 5 = 31$ .
2. Jump over the $1$ -st trap.Total damage: $\underline{0} + (10 + 1) + (11 + 1) + (5 + 1) = 29$ .
3. Jump over the $2$ -nd trap.Total damage: $5 + \underline{0} + (11 + 1) + (5 + 1) = 23$ .
4. Jump over the $3$ -rd trap.Total damage: $5 + 10 + \underline{0} + (5 + 1) = 21$ .
5. Jump over the $4$ -th trap.Total damage: $5 + 10 + 11 + \underline{0} = 26$ .

To get minimal damage it is needed to jump over the $3$ -rd trap, so the answer is $21$ .

In the third test case it is optimal to jump over the traps $1$ , $3$ , $4$ , $5$ , $7$ :

Total damage: $0 + (2 + 1) + 0 + 0 + 0 + (2 + 4) + 0 = 9$ .