human
Can you provide a Python code to print t integers given the following:
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.
assistant
#Here is Python code to print t integers.

''' Thruth can only be found at one place - THE CODE '''

''' Copyright 2018, SATYAM KUMAR'''
from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27)  # new thread will get stack of such size

printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
sys.setrecursionlimit(10000000)

class Operation:
    def __init__(self, name, function, function_on_equal, neutral_value=0):
        self.name = name
        self.f = function
        self.f_on_equal = function_on_equal
def add_multiple(x, count):
    return x * count
def min_multiple(x, count):
    return x
def max_multiple(x, count):
    return x
sum_operation = Operation("sum", sum, add_multiple, 0)
min_operation = Operation("min", min, min_multiple, 1e9)
max_operation = Operation("max", max, max_multiple, -1e9)
class SegmentTree:
    def __init__(self,
                 array,
                 operations=[sum_operation, min_operation, max_operation]):
        self.array = array
        if type(operations) != list:
            raise TypeError("operations must be a list")
        self.operations = {}
        for op in operations:
            self.operations[op.name] = op
        self.root = SegmentTreeNode(0, len(array) - 1, self)
    def query(self, start, end, operation_name):
        if self.operations.get(operation_name) == None:
            raise Exception("This operation is not available")
        return self.root._query(start, end, self.operations[operation_name])
    def summary(self):
        return self.root.values
    def update(self, position, value):
        self.root._update(position, value)
    def update_range(self, start, end, value):
        self.root._update_range(start, end, value)
    def __repr__(self):
        return self.root.__repr__()
class SegmentTreeNode:
    def...