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;
int a, b, q, c, d;
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
long long cal(long long x) {
  long long tmp1 = x / d, tmp2 = x % d;
  long long ans = tmp1 * (d - c) + max(tmp2 - c + 1, 0ll);
  return ans;
}
int main() {
  int t;
  cin >> t;
  for (int f = 0; f < t; f++) {
    int q;
    cin >> a >> b;
    cin >> q;
    c = max(a, b);
    d = (a * b) / gcd(a, b);
    for (int i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      cout << cal(r) - cal(l - 1) << endl;
    }
    cout << endl;
  }
  return 0;
}