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>
int func(int a, int b) {
  if (b == 0) return a;
  return func(b, a % b);
}
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int a, b, q;
    cin >> a >> b >> q;
    int lcm = (a * b) / func(a, b);
    int hm[lcm + 1];
    for (int i = 0; i <= lcm; i++) hm[i] = 0;
    for (int i = 1; i <= lcm; i++) {
      if ((i % a % b) != (i % b % a)) {
        hm[i] = hm[i - 1] + 1;
      } else
        hm[i] = hm[i - 1];
    }
    while (q--) {
      long long l;
      long long r;
      cin >> l >> r;
      if (r < a && r < b) {
        cout << 0 << endl;
        continue;
      }
      long long llim = (l / lcm) * lcm;
      long long ans = 0;
      if (llim != l) {
        llim += lcm;
        ans = ans + (hm[lcm] - (hm[l % lcm - 1]));
      }
      long long rlim = (r / lcm) * lcm;
      int diff = (rlim - llim) / lcm;
      ans = ans + (diff * hm[lcm]);
      diff = r - rlim;
      ans = ans + hm[diff];
      cout << ans << endl;
    }
  }
}
