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;
int main() {
  long long int t;
  cin >> t;
  for (long long int o = 0; o < t; o = o + 1) {
    long long int a, b, q;
    cin >> a >> b >> q;
    long long int c = a * b + 1;
    long long int d[c];
    d[0] = 0;
    for (long long int i = 1; i < c; i = i + 1) {
      d[i] = d[i - 1];
      if (((i % a) % b) != ((i % b) % a)) {
        d[i]++;
      }
    }
    c--;
    long long int l[q], r[q];
    long long int e, f;
    for (long long int i = 0; i < q; i = i + 1) {
      cin >> l[i] >> r[i];
      e = d[r[i] % c] + (r[i] / c) * d[c];
      l[i]--;
      f = d[l[i] % c] + (l[i] / c) * d[c];
      cout << e - f << " ";
    }
    cout << endl;
  }
}