#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long gcd(long long x, long long y) {
  if (x > y) swap(x, y);
  if (x == 0) return y;
  if (y % x == 0) return x;
  return gcd(y % x, x);
}
int main() {
  long long n, k, t;
  long long m, d, w, g;
  long long x, y, z;
  long long s, ans;
  bool v = true;
  ans = 0;
  scanf("%lld", &t);
  for (long long tt = 0; tt < (long long)(t); tt++) {
    scanf("%lld %lld %lld", &m, &d, &w);
    if ((m == 1) || (d == 1)) {
      printf("0\n");
      continue;
    }
    g = gcd(d - 1, w);
    k = w / g;
    x = min(m, d);
    y = x % k;
    x = x / k;
    s = (x + 1) * x;
    s /= 2;
    ans = s * y;
    s = x * (x - 1);
    s /= 2;
    ans += (s * (k - y));
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

Berland year consists of m months with d days each. Months are numbered from 1 to m. Berland week consists of w days. The first day of the year is also the first day of the week. Note that the last week of the year might be shorter than w days.

A pair (x, y) such that x < y is ambiguous if day x of month y is the same day of the week as day y of month x.

Count the number of ambiguous pairs.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

Each of the next t lines contains three integers m, d and w (1 ≤ m, d, w ≤ 10^9) — the number of months in a year, the number of days in a month and the number of days in a week.

Output

Print t integers — for each testcase output the number of pairs (x, y) such that x < y and day x of month y is the same day of the week as day y of month x.

Example

Input


5
6 7 4
10 7 12
12 30 7
1 1 1
3247834 10298779 625324


Output


6
9
5
0
116461800

Note

Here are the pairs for the first test case: 

<image>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long gcd(long long x, long long y) {
  if (x > y) swap(x, y);
  if (x == 0) return y;
  if (y % x == 0) return x;
  return gcd(y % x, x);
}
int main() {
  long long n, k, t;
  long long m, d, w, g;
  long long x, y, z;
  long long s, ans;
  bool v = true;
  ans = 0;
  scanf("%lld", &t);
  for (long long tt = 0; tt < (long long)(t); tt++) {
    scanf("%lld %lld %lld", &m, &d, &w);
    if ((m == 1) || (d == 1)) {
      printf("0\n");
      continue;
    }
    g = gcd(d - 1, w);
    k = w / g;
    x = min(m, d);
    y = x % k;
    x = x / k;
    s = (x + 1) * x;
    s /= 2;
    ans = s * y;
    s = x * (x - 1);
    s /= 2;
    ans += (s * (k - y));
    printf("%lld\n", ans);
  }
  return 0;
}

```