Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> marr(400000);
long long a, b, q;
long long summ(long long bruh) {
  long long res = (bruh / (a * b)) * marr[a * b] + marr[bruh % (a * b)];
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int notc;
  cin >> notc;
  while (notc--) {
    cin >> a >> b >> q;
    for (int i = 0; i < a * b; i++) {
      marr[i] = 0;
    }
    for (int i = 0; i < a * b; i++)
      marr[i + 1] = marr[i] + (((i % a) % b) != ((i % b) % a));
    while (q--) {
      long long li, ri;
      cin >> li >> ri;
      cout << (summ(ri + 1) - summ(li)) << " ";
    }
    cout << endl;
  }
  return 0;
}