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;
struct pair_hash {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2> &p) const {
    auto h1 = std::hash<T1>{}(p.first);
    auto h2 = std::hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
using vote = pair<long long int, long long int>;
using Unordered_map = unordered_map<vote, long long int, pair_hash>;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int a, b, q;
    cin >> a >> b >> q;
    long long int m = a * b;
    long long int rem[50000] = {};
    for (long long int i = 1; i < m; i++) {
      rem[i] = rem[i - 1];
      long long int x = ((i % a) % b);
      long long int y = ((i % b) % a);
      if (x != y) rem[i]++;
    }
    while (q--) {
      long long int l, r;
      cin >> l >> r;
      long long int c = 0;
      if (a == b || a % b == 0 || b % a == 0 || a == 1 || b == 1) {
        cout << 0 << " ";
        continue;
      }
      c = (r / m) * rem[m - 1];
      c += rem[r % m];
      c -= (l / m) * rem[m - 1];
      c -= rem[(l % m) - 1];
      cout << c << " ";
    }
    cout << endl;
  }
}
