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 T;
long long a, b, q, l, r;
int main() {
  cin >> T;
  while (T--) {
    cin >> a >> b >> q;
    bool tab[a * b];
    long long sum = 0LL;
    long long sus[a * b + 1];
    sus[0] = 0;
    for (int i = 0; i < a * b; i++) {
      tab[i] = ((i % b) % a != (i % a) % b);
      if (tab[i]) sus[i]++;
      sus[i + 1] = sus[i];
      if (tab[i]) sum++;
    }
    long long res;
    while (q--) {
      res = 0;
      cin >> l >> r;
      if (l % (a * b) > 0) res -= sus[l % (a * b) - 1];
      l -= l % (a * b);
      res += sus[r % (a * b)];
      r -= r % (a * b);
      res += sum * ((r - l) / (a * b));
      cout << res << " \n"[q == 0];
    }
  }
  return 0;
}
