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 count(long long d, long long ab, long long arr[]) {
  long long v = (d / ab) * arr[ab - 1];
  long long q = d % ab;
  v += arr[q];
  return v;
}
int main(int argc, char const *argv[]) {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long arr[40000] = {0};
    for (long long i = 1; i < a * b; ++i) {
      long long v1 = (i % a) % b;
      long long v2 = (i % b) % a;
      if (v1 != v2) {
        arr[i]++;
      }
      arr[i] += arr[i - 1];
    }
    while (q--) {
      long long l, r;
      cin >> l >> r;
      cout << count(r, a * b, arr) - count(l - 1, a * b, arr) << " ";
    }
    cout << endl;
  }
  return 0;
}