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() {
  int t;
  cin >> t;
  while (t--) {
    long long int a, b, q;
    cin >> a >> b >> q;
    long long int s = max(a, b);
    long long int m = (a * b);
    std::vector<long long int> sol;
    while (q--) {
      long long int l, r, num = 0;
      cin >> l >> r;
      if (l <= m && r <= m) {
        for (int i = l; i <= r; i++) {
          if ((i % a) % b != (i % b) % a) num++;
        }
        sol.push_back(num);
      } else if (l <= m && r > m) {
        for (int i = l; i <= m; i++) {
          if ((i % a) % b != (i % b) % a) num++;
        }
        long long int q2 = (r - m) / m;
        long long int s1 = q2 * s;
        num = num + ((r - m) - s1);
        sol.push_back(num);
      } else {
        long long int q1 = l / m;
        long long int q2 = r / m;
        long long int total = (q2 - q1);
        long long int sum = total * s;
        sol.push_back((r - l + 1) - sum);
      }
    }
    for (int i = 0; i < sol.size(); i++) cout << sol[i] << " ";
    cout << endl;
  }
  return 0;
}
