Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void solve() {
  long long a, b, q;
  cin >> a >> b >> q;
  long long mx = max(a, b);
  long long lcm = a / gcd(a, b) * b;
  long long num = lcm - mx;
  long long l, r, d1, d2;
  long long ans;
  long long sub;
  while (q--) {
    cin >> l >> r;
    d1 = l / lcm;
    d2 = r / lcm;
    ans = (d2 - d1) * num;
    ans = ans - max(0ll, l - d1 * lcm - mx) + max(0ll, r + 1 - d2 * lcm - mx);
    cout << ans << " ";
  }
  cout << endl;
}
int32_t main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) solve();
}