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;
const long long int N = 200010;
const long long mod = 1000000007;
const long long inf = 1000000000000000000;
const long double pi = 3.14159265358979323846264338;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    long long int _lcm = lcm(a, b);
    long long int q;
    cin >> q;
    while (q--) {
      long long int l, r;
      cin >> l >> r;
      long long int s = 0, e = 0;
      if (r < b || _lcm == b)
        cout << "0"
             << " ";
      else if (l >= b && r >= b)
        s = l, e = r;
      else
        s = b, e = r;
      if (s == 0 && e == 0) continue;
      long long int it = _lcm;
      if (_lcm < s) {
        if (s % _lcm == 0)
          _lcm = (s / _lcm) * _lcm;
        else
          _lcm = ((s / _lcm) + 1) * _lcm;
      }
      long long int ans = (e - s + 1);
      for (long long int j = _lcm; j <= e; j += it) {
        for (long long int i = j; i <= min({j + b - 1, e}); i++) {
          if ((i % a) % b == (i % b) % a) {
            ans--;
          }
        }
      }
      cout << ans << " ";
    }
    cout << "\n";
  }
  return 0;
}
