Problem: 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 ≤ t ≤ 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 ≤ a, b, k ≤ 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.

Example

Input


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


Output


8
198
-17
2999999997
0
1

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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class FrogJumping {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		// Problem http://codeforces.com/problemset/problem/1077/A
		// For more solutions check https://github.com/jontiboss
		Reader reader = new Reader();
		int firstJump;
		int secondJump;
		long jumps;
		int queries = reader.nextInt();
		for(int i =0;i<queries;i++) {
			firstJump = reader.nextInt();
			secondJump = reader.nextInt();
			jumps = reader.nextInt();
			//If the jump is even the distance we will end up at is equal to (firstJump-second jump)*jumps/2, if its odd we could just add one to the jumps and add one second Jump
			// to the sum.
			// the idea will be as follow, for three jumps we will have A-B+A=2A-b and this is the same as A-B+A-B+B
			// for four jumps we will have A-B+A-B = 2A-2B = 2(A-B) the factor of two is equal to 4/2
			System.out.println((jumps&1)==0 ? (firstJump-secondJump)*(jumps>>1): (firstJump-secondJump)*((jumps+1)>>1)+secondJump);
		}
	}
	static class Reader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public Reader() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				try
				{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException e) 
				{ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 

		int nextInt() 
		{ 
			return Integer.parseInt(next()); 
		} 
		long nextLong() 
		{ 
			return Long.parseLong(next()); 
		} 
	}
}