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;
int32_t main() {
  long long t;
  cin >> t;
  for (long long q = 0; q < t; q++) {
    long long a, b, k;
    cin >> a >> b >> k;
    vector<long long> x(a * b);
    for (long long i = 0; i < a * b; i++) {
      x[i] = (i % a % b != i % b % a);
    }
    long long all = 0;
    for (long long i = 0; i < a * b; i++) {
      all += x[i];
    }
    vector<long long> sum_pref(a * b + 1, 0);
    for (long long i = 0; i < a * b; i++) {
      sum_pref[i + 1] = sum_pref[i] + x[i];
    }
    for (long long i = 0; i < k; i++) {
      long long l, r;
      cin >> l >> r;
      r++;
      long long ans = (r / (a * b) - l / (a * b)) * all;
      ans += sum_pref[r % (a * b)];
      ans -= sum_pref[l % (a * b)];
      cout << ans << " ";
    }
    cout << endl;
  }
  return 0;
}
