You are given two integers a and b, and q queries. The i-th query consists of two numbers l_i and r_i, and the answer to it is the number of integers x such that l_i ≤ x ≤ r_i, and ((x mod a) mod b) ≠ ((x mod b) mod a). Calculate the answer for each query.

Recall that y mod z is the remainder of the division of y by z. For example, 5 mod 3 = 2, 7 mod 8 = 7, 9 mod 4 = 1, 9 mod 9 = 0.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then the test cases follow.

The first line of each test case contains three integers a, b and q (1 ≤ a, b ≤ 200; 1 ≤ q ≤ 500).

Then q lines follow, each containing two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^{18}) for the corresponding query.

Output

For each test case, print q integers — the answers to the queries of this test case in the order they appear.

Example

Input


2
4 6 5
1 1
1 3
1 5
1 7
1 9
7 10 2
7 8
100 200


Output


0 0 0 2 4 
0 91 
#include <bits/stdc++.h>
using namespace std;
using ll = signed long long int;
const int N = 1e5;
ll a, b, ab, q, l, r, res, x[N];
void solve() {
  cin >> a >> b >> q;
  ab = a * b;
  for (int i = 1; i <= ab; i++) {
    x[i] = x[i - 1];
    if (((i % a) % b) != ((i % b) % a)) {
      x[i]++;
    }
  }
  while (q--) {
    cin >> l >> r;
    res = 0LL;
    if (r - l + 1LL >= ab) {
      if (l % ab != 1LL) {
        res += x[ab] - x[(l % ab) - 1];
        l += ab - (l % ab) + 1LL;
      }
      if (r % ab != 1LL) {
        res += x[r % ab];
        r -= r % ab;
      }
      res += max(r - l + 1, 0LL) / ab * x[ab];
    } else {
      if (r % ab >= (l % ab) - 1) {
        res = x[r % ab] - x[(l % ab) - 1];
      } else {
        res = x[r % ab] + x[ab] - x[(l % ab) - 1];
      }
    }
    cout << res << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cout << fixed;
  cout.precision(12);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
