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_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t1;
  cin >> t1;
  while (t1--) {
    long long int a, b, q;
    cin >> a >> b >> q;
    while (q--) {
      long long int l, r;
      cin >> l >> r;
      long long int out = 0;
      long long int a1 = min(a, b);
      long long int b1 = max(a, b);
      if (a1 % b1 == 0)
        cout << "0"
             << " ";
      else {
        if (l <= b1) {
          if (r < b1)
            cout << 0 << " ";
          else {
            out = abs(r - b1);
            out = out / b1 * b1 + out % b1;
            cout << out + 1 << " ";
          }
        } else {
          if (r < b1)
            cout << 0 << " ";
          else {
            out = abs(r - l - b1);
            out = out / b1 * b1 + out % b1;
            cout << out + 1 << " ";
          }
        }
      }
    }
    cout << "\n";
  }
}
