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;
int main() {
  int t;
  cin >> t;
  for (int it = 0; it < t; it++) {
    int a, b, q;
    cin >> a >> b >> q;
    int interval = a * b;
    vector<int> solution_set;
    int count = 0;
    solution_set.push_back(0);
    for (int x = 1; x <= interval; x++) {
      if ((x % a) % b != (x % b) % a) count++;
      solution_set.push_back(count);
    }
    long long left_count, right_count, remainder;
    vector<long long> answer;
    for (int i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      l--;
      left_count = (l / interval) * solution_set.back();
      remainder = l % interval;
      left_count += solution_set[remainder];
      right_count = (r / interval) * solution_set.back();
      remainder = r % interval;
      right_count += solution_set[remainder];
      answer.push_back(right_count - left_count);
    }
    for (int x : answer) cout << x << " ";
    cout << endl;
  }
}
