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;
int64_t fu(int64_t l, int64_t val, int64_t bigger) {
  int64_t total = l + 1;
  int64_t ans = 0;
  int64_t count = l / val;
  if (count > 0) {
    ans = ans + (bigger) * (count - 1);
  }
  int64_t tmp = 1;
  int64_t our = min(l - val * count, bigger - 1);
  tmp += our;
  ans += tmp;
  return total - ans;
}
int64_t gcd(int64_t a, int64_t b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int64_t t;
  cin >> t;
  while (t--) {
    int64_t a, b, q;
    cin >> a >> b >> q;
    int64_t aa, bb;
    int64_t lcm = a * b / (gcd(a, b));
    int64_t maax = max(a, b);
    while (q--) {
      cin >> aa >> bb;
      int64_t one = fu(bb, lcm, maax);
      int64_t to = fu(aa - 1, lcm, maax);
      cout << one - to << " ";
    }
    cout << endl;
  }
}
