Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int cnt(long long int n, int* arr, int* csum, long long int a,
                  long long int b) {
  long long int a1 = (n / (a * b)) * csum[a * b - 1];
  long long int q = n % (a * b);
  long long int a2 = csum[q];
  return a1 + a2;
}
int main() {
  long long int t, a, b, q, l, r;
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    int arr[a * b];
    int csum[a * b];
    for (int i = 0; i < a * b; i++) {
      if ((i % a) % b != (i % b) % a) {
        arr[i] = 1;
      } else {
        arr[i] = 0;
      }
      csum[i] = arr[i] + csum[i - 1];
    }
    csum[0] = arr[0];
    for (int i = 1; i < a * b; i++) {
      csum[i] = arr[i] + csum[i - 1];
    }
    while (q--) {
      cin >> l >> r;
      long long int ans = cnt(r, arr, csum, a, b) - cnt(l - 1, arr, csum, a, b);
      cout << ans << " ";
    }
    cout << endl;
  }
  return 0;
}