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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lec(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    if (a < b) {
      swap(a, b);
    }
    long long lcm = lec(a, b);
    for (int i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      if (a == b) {
        cout << "0 ";
        continue;
      }
      long long res = 0;
      long long l1 = (l / lcm);
      long long r1 = (r / lcm);
      if (r1 == l1) {
        if (l1 * lcm + a > l) {
          res += min(l1 * lcm + a - l, r - l + 1);
        }
      } else {
        res += (r1 - l1) * a;
        if (l > l1 * lcm) {
          res -= min(l - l1 * lcm, a);
        }
        if (r >= r1 * lcm) {
          res += min(a, r - r1 * lcm + 1);
        }
      }
      long long k = r - l + 1 - res;
      cout << k << " ";
    }
    cout << "\n";
  }
  return 0;
}