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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T, class S>
ostream &operator<<(ostream &os, const pair<T, S> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class S>
ostream &operator<<(ostream &os, const unordered_map<T, S> &v) {
  for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' ';
  return os;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  for (int ti = 1; ti <= t; ti++) {
    long long a, b, q, a11, b11;
    cin >> a11 >> b11 >> q;
    a = min(a11, b11);
    b = max(a11, b11);
    for (int i = int(0); i <= int(q - 1); i++) {
      long long l, r;
      cin >> l >> r;
      long long a1 = l % (a * b), a2 = r % (a * b), ans = 0;
      long long c1 = l / (a * b), c2 = r / (a * b);
      if (c2 == c1) {
        if (b >= a1) {
          if (b <= a2) {
            ans += b - a1;
          } else {
            ans += r - l + 1;
          }
        }
      } else {
        ans += (c2 - c1 - 1) * b;
        if (b - 1 >= a1) ans += b - a1;
        if (a2 <= b - 1)
          ans += a2;
        else
          ans += b;
      }
      cout << (r - l + 1 - ans) << " ";
    }
    cout << "\n";
  }
  return 0;
}
