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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    vector<long long> ans;
    long long lcmval = lcm(a, b);
    while (q--) {
      long long l, r;
      cin >> l >> r;
      if (lcmval == 1) {
        ans.push_back(0);
      } else {
        long long x = (r - l) / lcmval;
        long long y = (r - l) % lcmval;
        int ct = 0;
        for (long long i = l; i <= min(lcmval + l - 1, r); i++) {
          if ((i % a) % b != (i % b) % a) {
            ct++;
          }
        }
        if (x == 0) {
          ans.push_back(ct);
        } else {
          for (long long i = r - y; i <= r; i++) {
            if ((i % a) % b != (i % b) % a) {
              ct++;
            }
          }
          ans.push_back(ct);
        }
      }
    }
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
}
