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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const pair<T, U> &_p) {
  _out << _p.first << ' ' << _p.second;
  return _out;
}
template <typename T, typename U>
inline istream &operator>>(istream &_in, pair<T, U> &_p) {
  _in >> _p.first >> _p.second;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
  if (_v.empty()) {
    return _out;
  }
  _out << _v.front();
  for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) {
    _in >> _i;
  }
  return _in;
}
const long long MAXN = 1e5;
const long long INF = 1e9;
const long long MOD = 1e9 + 7;
long long process(long long a, long long b) { return min(a, b); }
class BGermes {
 public:
  void solve(std::istream &in, std::ostream &out) {
    long long q;
    in >> q;
    for (long long i = 0; i < q; i++) {
      long long a, b, k;
      in >> a >> b >> k;
      out << a * ((k + 1) / 2) - b * (k / 2) << endl;
    }
  }
};
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  BGermes solver;
  istream &in(cin);
  ostream &out(cout);
  solver.solve(in, out);
  return 0;
}