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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    long long a, b, q;
    cin >> a >> b >> q;
    vector<long long> prefixsum;
    prefixsum.push_back(0);
    for (long long j = 1; j <= a * b; j++) {
      if ((j % a) % b != (j % b) % a) {
        prefixsum.push_back(prefixsum[j - 1] + 1);
      } else {
        prefixsum.push_back(prefixsum[j - 1]);
      }
    }
    for (long long j = 0; j < q; j++) {
      long long l, r;
      cin >> l >> r;
      if (r <= a * b) {
        cout << prefixsum[r] - prefixsum[l - 1] << " ";
      } else {
        long long right =
            (r / (a * b)) * prefixsum[a * b] + prefixsum[r % (a * b)];
        long long left = ((l - 1) / (a * b)) * prefixsum[a * b] +
                         prefixsum[(l - 1) % (a * b)];
        ;
        cout << right - left << " ";
      }
    }
    cout << "\n";
  }
  return 0;
}
