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() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int tc;
  cin >> tc;
  while (tc--) {
    int a, b, q, n;
    cin >> a >> b >> q;
    n = a * b;
    vector<int> query(n, 0);
    int counter = 0;
    for (int i = 0; i < n; i++) {
      if (((i % b) % a) != ((i % a) % b)) {
        ++counter;
      }
      query[i] = counter;
    }
    for (int i = 0; i < q; i++) {
      long long x, y;
      cin >> x >> y;
      long long q1, q2;
      x--;
      q1 = x / n * counter + query[(x) % n];
      q2 = y / n * counter + query[y % n];
      cout << q2 - q1 << endl;
    }
  }
  return 0;
}
