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.
Input
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 $2 \cdot 10^5$.
Output
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.
Note
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:
- Do not jump over any trap.Total damage:
$5 + 10 + 11 + 5 = 31$. - Jump over the
$1$-st trap.Total damage:$\underline{0} + (10 + 1) + (11 + 1) + (5 + 1) = 29$. - Jump over the
$2$-nd trap.Total damage:$5 + \underline{0} + (11 + 1) + (5 + 1) = 23$. - Jump over the
$3$-rd trap.Total damage:$5 + 10 + \underline{0} + (5 + 1) = 21$. - 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: $0 + (2 + 1) + 0 + 0 + 0 + (2 + 4) + 0 = 9$.
样例
5
4 4
8 7 1 4
4 1
5 10 11 5
7 5
8 2 5 15 11 2 8
6 3
1 2 3 4 5 6
1 1
7
0
21
9
6
0
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