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 gcd(int a, int b) {
  if (a % b == 0) {
    return b;
  }
  return gcd(b, a % b);
}
int main() {
  int test = 1;
  cin >> test;
  for (int t = 0; t < test; t++) {
    int a, b, q;
    cin >> a >> b >> q;
    int maxi = max(a, b);
    int lcm = (a * b) / gcd(a, b);
    vector<int> values(lcm + 1, 0);
    for (int i = 1; i <= lcm; i++) {
      if (i % a % b == i % b % a) {
        values[i] = values[i - 1] + 1;
      } else {
        values[i] = values[i - 1];
      }
    }
    for (int i = 0; i < q; i++) {
      long long int l, r;
      cin >> l >> r;
      int a1 = (r / lcm) * values[lcm] + values[r % lcm];
      int a2 = ((l - 1) / lcm) * values[lcm] + values[(l - 1) % lcm];
      cout << r - l + 1 - (a1 - a2) << " ";
    }
    cout << endl;
  }
  return 0;
}
