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);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  long long num1, num2, a, b, q, l, r;
  while (t--) {
    cin >> num1 >> num2 >> q;
    a = min(num1, num2);
    b = max(num1, num2);
    vector<int> ans;
    long long lcm = (a * b) / gcd(a, b);
    while (q--) {
      cin >> l >> r;
      if (r < b) {
        ans.push_back(0);
      } else {
        long long count = 0;
        if (l < b) {
          l = b;
        }
        int lo = ceil(float(l) / float(lcm));
        int hi = ceil(float(r) / float(lcm));
        count += (r - l + 1) - ((hi - lo) * b);
        ans.push_back(count);
      }
    }
    for (auto s : ans) {
      cout << s << " ";
    }
    cout << "\n";
  }
}
