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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  map<long long int, long long int> m;
  long long int a, b, q;
  cin >> a >> b >> q;
  for (long long int i = 0; i < (a * b) / gcd(a, b) + 1; i++) {
    if (i == 0) continue;
    if ((i % a) % b == (i % b) % a)
      m[i] = m[i - 1];
    else
      m[i] = m[i - 1] + 1;
  }
  long long int lcm = (a * b) / gcd(a, b);
  for (long long int i = 0; i < q; i++) {
    long long int l, r;
    cin >> l >> r;
    long long int k = (r / lcm) * m[lcm] + m[r % lcm];
    if (l != 1) {
      l--;
      k -= (l / lcm) * m[lcm] + m[l % lcm];
    }
    cout << k << " ";
  }
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t1;
  t1 = 1;
  cin >> t1;
  while (t1--) {
    solve();
  }
  return 0;
}