D. Running Miles
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

There is a street with $n$ sights, with sight number $i$ being $i$ miles from the beginning of the street. Sight number $i$ has beauty $b_i$. You want to start your morning jog $l$ miles and end it $r$ miles from the beginning of the street. By the time you run, you will see sights you run by (including sights at $l$ and $r$ miles from the start). You are interested in the $3$ most beautiful sights along your jog, but every mile you run, you get more and more tired.

So choose $l$ and $r$, such that there are at least $3$ sights you run by, and the sum of beauties of the $3$ most beautiful sights minus the distance in miles you have to run is maximized. More formally, choose $l$ and $r$, such that $b_{i_1} + b_{i_2} + b_{i_3} - (r - l)$ is maximum possible, where $i_1, i_2, i_3$ are the indices of the three maximum elements in range $[l, r]$.

Input

The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 10^5$).

The second line of each test case contains $n$ integers $b_i$ ($1 \leq b_i \leq 10^8$) — beauties of sights $i$ miles from the beginning of the street.

It's guaranteed that the sum of all $n$ does not exceed $10^5$.

Output

For each test case output a single integer equal to the maximum value $b_{i_1} + b_{i_2} + b_{i_3} - (r - l)$ for some running range $[l, r]$.

Example
Input
455 1 4 2 341 1 1 169 8 7 6 5 47100000000 1 100000000 1 100000000 1 100000000
Output
8
1
22
299999996

Note

In the first example, we can choose $l$ and $r$ to be $1$ and $5$. So we visit all the sights and the three sights with the maximum beauty are the sights with indices $1$, $3$, and $5$ with beauties $5$, $4$, and $3$, respectively. So the total value is $5 + 4 + 3 - (5 - 1) = 8$.

In the second example, the range $[l, r]$ can be $[1, 3]$ or $[2, 4]$, the total value is $1 + 1 + 1 - (3 - 1) = 1$.