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 i = 0; i < t; i++) {
    int a, b, q;
    cin >> a >> b >> q;
    int r1 = a * b;
    int c[r1];
    c[0] = 0;
    for (int j = 1; j < r1; j++) {
      if (((j % a) % b) != ((j % b) % a)) {
        c[j] = c[j - 1] + 1;
      } else
        c[j] = c[j - 1];
    }
    for (int j = 0; j < q; j++) {
      long long int l, r;
      cin >> l >> r;
      long long int count = 0;
      long long int x = (l - 1) / r1;
      long long int y = (l - 1) % r1;
      long long int x1 = r / r1;
      long long int y1 = r % r1;
      long long int z = (x * c[r1 - 1]) + c[y];
      long long int z1 = (x1 * c[r1 - 1]) + c[y1];
      count = z1 - z;
      cout << count << " ";
    }
    cout << endl;
  }
  return 0;
}
