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--) {
    int a, b, q;
    cin >> a >> b >> q;
    vector<int> v(a * b + 1, 0);
    for (int i = 1; i <= a * b; i++) {
      if ((i % a) % b != (i % b) % a) {
        v[i] = v[i - 1] + 1;
      } else
        v[i] = v[i - 1];
    }
    while (q--) {
      long long l, r;
      cin >> l >> r;
      long long l_cycle = l / (a * b);
      long long r_cycle = r / (a * b);
      long long ans = (r_cycle - l_cycle) * v[a * b];
      ans = ans + v[r % (a * b)];
      if (l % (a * b) != 0) ans = ans - v[(l % (a * b)) - 1];
      cout << ans << endl;
    }
  }
  return 0;
}
