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;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(16);
  int t;
  cin >> t;
  for (int ii = 0; ii < t; ii++) {
    long long int a, b, q;
    cin >> a >> b >> q;
    long long int g = lcm(a, b);
    for (int j = 0; j < q; j++) {
      long long int l, r;
      cin >> l >> r;
      long long int m = (r - l) / g;
      long long int ans = 0;
      long long int f, last;
      f = (l / g) * g;
      last = (r / g) * g;
      if (a > b) {
        if (f + a - l > 0) ans += (f + a - l);
        ans += ((last - f) / g - 1) * a;
        if (r - last + 1 >= a) {
          ans += a;
        } else {
          ans += r - last + 1;
        }
        ans = (r - l + 1) - ans;
      } else {
        if (f + b - l > 0) ans += (f + b - l);
        ans += ((last - f) / g - 1) * b;
        if (r - last + 1 >= b) {
          ans += b;
        } else {
          ans += r - last + 1;
        }
        ans = (r - l + 1) - ans;
      }
      cout << ans << " ";
    }
    cout << endl;
  }
  return 0;
}
