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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int a, b, q;
    cin >> a >> b >> q;
    int rem = a * b;
    vector<int> pref(rem + 1);
    pref[0] = 0;
    for (int x = 1; x <= rem; x++) {
      pref[x] = pref[x - 1];
      if ((x % a) % b != (x % b) % a) pref[x]++;
    }
    vector<long long> res;
    for (int i = 0; i < q; i++) {
      long long L, R;
      cin >> L >> R;
      res.push_back(((R / rem) * 1LL * pref[rem] + pref[R % rem]) -
                    (((L - 1) / rem) * 1LL * pref[rem] + pref[(L - 1) % rem]));
    }
    for (long long i : res) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}
