# Frog Jumping

## 题意翻译

###### 题目描述
对于给定的$a,b,k$，初始状态$x=0$，第$i$次操作满足以下规则：
- $i$为奇数，$x=x+a$
- $i$为偶数，$x=x-b$
求$k$次操作后$x$的值
###### 输入格式
**题目有多组数据**
第一行一个整数$t(1\leq t\leq1000)$表示数据组数
接下来$t$行，每行三个整数$a,b,k(1\leq a,b,k\leq10^9)$表示每组数据
###### 输出格式
输出共$t$行，第$i$行表示第$i$组数据的答案

## 题目描述

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.

## 输入输出格式

### 输入格式

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.

### 输出格式

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

## 输入输出样例

### 输入样例 #1

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

### 输出样例 #1

```
8
198
-17
2999999997
0
1
```

## 说明

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 $ .