A frog is currently at the point $0$ on a coordinate axis $Ox$. It jumps by the following algorithm: the first jump is $a$ units to the right, the second jump is $b$ units to the left, the third jump is $a$ units to the right, the fourth jump is $b$ units to the left, and so on.

Formally:

• if the frog has jumped an even number of times (before the current jump), it jumps from its current position $x$ to position $x+a$;
• otherwise it jumps from its current position $x$ to position $x-b$.

Your task is to calculate the position of the frog after $k$ jumps.

But... One more thing. You are watching $t$ different frogs so you have to answer $t$ independent queries.

Input

The first line of the input contains one integer $t$ ($1 \le t \le 1000$) — the number of queries.

Each of the next $t$ lines contain queries (one query per line).

The query is described as three space-separated integers $a, b, k$ ($1 \le a, b, k \le 10^9$) — the lengths of two types of jumps and the number of jumps, respectively.

Output

Print $t$ integers. The $i$-th integer should be the answer for the $i$-th query.

Note

In the first query frog jumps $5$ to the right, $2$ to the left and $5$ to the right so the answer is $5 - 2 + 5 = 8$.

In the second query frog jumps $100$ to the right, $1$ to the left, $100$ to the right and $1$ to the left so the answer is $100 - 1 + 100 - 1 = 198$.

In the third query the answer is $1 - 10 + 1 - 10 + 1 = -17$.

In the fourth query the answer is $10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997$.

In the fifth query all frog's jumps are neutralized by each other so the answer is $0$.

The sixth query is the same as the fifth but without the last jump so the answer is $1$.

Samples

6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999

8
198
-17
2999999997
0
1