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;
    vector<pair<long long, long long> > queries;
    long long largest = 0;
    long long count = 0;
    for (int i = 1; i <= 100; i++) {
      if (((i % a) % b) != ((i % b) % a)) {
        count++;
      }
    }
    for (int i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      long long ans = 0;
      if (r - l > 100) {
        long long tmpL = l;
        while (tmpL % 100 != 0) {
          if (((tmpL % a) % b) != ((tmpL % b) % a)) {
            ans++;
          }
          tmpL++;
        }
        if (((tmpL % a) % b) != ((tmpL % b) % a)) {
          ans++;
        }
        long long remaining = r - tmpL;
        long long leftover = remaining % 100;
        long long times = floor(remaining / 100);
        ans += (times * count);
        for (int j = 1; j <= leftover; j++) {
          if (((j % a) % b) != ((j % b) % a)) {
            ans++;
          }
        }
      } else {
        for (int j = l; j <= r; j++) {
          if (((j % a) % b) != ((j % b) % a)) {
            ans++;
          }
        }
      }
      cout << ans << " ";
    }
    cout << "\n";
  }
}
