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::sync_with_stdio(0);
  cin.tie(0);
  int nTest;
  cin >> nTest;
  for (int _ = 0; _ < nTest; _++) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long prefix[100001] = {0};
    for (int i = 1; i <= 100000; i++) {
      if (((i % a) % b) != ((i % b) % a)) {
        prefix[i]++;
      }
      prefix[i] += prefix[i - 1];
    }
    for (int i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      long long ans = 0;
      long long lTime = l / 100000;
      long long lRemainder = l % 100000;
      long long lAmount = 0;
      long long rTime = r / 100000;
      long long rRemainder = r % 100000;
      long long rAmount = 0;
      lRemainder = lRemainder > 0 ? lRemainder - 1 : lRemainder;
      lAmount = (lTime * prefix[100000]);
      lAmount += (prefix[lRemainder]);
      rAmount = (rTime * prefix[100000]);
      rAmount += (prefix[rRemainder]);
      ans = rAmount - lAmount;
      cout << ans << " ";
    }
    cout << "\n";
  }
}
