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;
void boostIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int t;
string str;
long long gcd(long long a, long long b) {
  while (a != 0) {
    b %= a;
    swap(a, b);
  }
  return b;
}
int main() {
  boostIO();
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long lcm = (a * b) / gcd(a, b);
    vector<long long> v;
    while (q--) {
      long long l, r;
      cin >> l >> r;
      long long ans = r - l + 1;
      ans -= max(a, b) * (r / lcm - l / lcm);
      ans -= l % lcm < max(a, b) ? max(a, b) - l % lcm : 0;
      ans += r % lcm < max(a, b) ? max(a, b) - r % lcm - 1 : 0;
      v.push_back(ans);
    }
    for (int u : v) cout << u << ' ';
    cout << '\n';
  }
  return 0;
}
