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() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b, q, l, r;
    cin >> a >> b >> q;
    long long int ab = a * b;
    vector<long long int> v(ab, 0);
    long long int cnt = 0;
    for (int i = 1; i <= ab; i++) {
      if ((i % a) % b != (i % b) % a) {
        cnt++;
      }
      v[i - 1] = cnt;
    }
    while (q--) {
      cin >> l >> r;
      long long int c, d, ans = 0;
      c = r % ab;
      d = (l - 1) % ab;
      ans = (r / ab) * v[ab - 1] - ((l - 1) / ab) * v[ab - 1];
      if (c != 0) ans += v[r % ab - 1];
      if (d != 0) ans -= v[(l - 1) % ab - 1];
      cout << ans << " ";
    }
    cout << endl;
  }
}
