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 LL = long long;
using namespace std;
int t, n;
int gcd(int a, int b) {
  if (a == 0) {
    return b;
  }
  int d = gcd(b % a, a);
  return d;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  for (LL ii = 0; ii < t; ii++) {
    LL a, b, q;
    cin >> a >> b >> q;
    LL l, r;
    for (LL i = 0; i < q; i++) {
      cin >> l >> r;
      LL full_cnt = 0;
      if (r - l + 1 <= 200) {
        for (LL j = l; j <= r; ++j) {
          if ((j % a) % b == (j % b) % a) full_cnt++;
        }
        cout << r - l + 1ll - full_cnt << " ";
        continue;
      }
      LL min_lcm_left = a * b / gcd(a, b),
         lcm_right = min_lcm_left + max(a, b) - 1;
      LL min_k_left = l / min_lcm_left, max_k_left = r / min_lcm_left;
      LL min_k_left_full = min_k_left;
      LL max_k_left_full = max_k_left;
      if (min_k_left * min_lcm_left < l) min_k_left_full++;
      if (max_k_left * min_lcm_left + max(a, b) > r) max_k_left_full--;
      if (max_k_left_full >= min_k_left_full)
        full_cnt = (max_k_left_full - min_k_left_full + 1) * max(a, b);
      if (min_k_left != min_k_left_full)
        for (LL j = max(l, min_k_left * min_lcm_left);
             j < min(r + 1, min_k_left * min_lcm_left + max(a, b)); ++j) {
          if (j >= l && j <= r && (j % a) % b == (j % b) % a) full_cnt++;
        }
      if (max_k_left != max_k_left_full && max_k_left_full >= 0)
        for (LL j = max(l, max_k_left * min_lcm_left);
             j < min(r + 1, max_k_left * min_lcm_left + max(a, b)); ++j) {
          if (j >= l && j <= r && (j % a) % b == (j % b) % a) full_cnt++;
        }
      cout << r - l + 1ll - full_cnt << " ";
    }
    cout << endl;
  }
  return 0;
}