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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long a, b, q, l, r;
long long temp;
long long cal(long long x) {
  long long ans = 0;
  long long num = x / temp;
  if (num == 0) return 0;
  ans += num * b;
  if (num * temp + b - 1 > x) ans -= (num * temp + b - 1 - x);
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    if (a > b) {
      long long temp = a;
      a = b;
      b = temp;
    }
    temp = lcm(a, b);
    for (int j = 1; j <= q; j++) {
      cin >> l >> r;
      long long ans = r - l + 1;
      if (l <= b) {
        if (r < b) {
          ans -= r - l + 1;
        } else if (r == b) {
          ans -= r - l;
        } else
          ans -= b - l;
      }
      ans -= (cal(r) - cal(l - 1));
      if (j == 1)
        cout << ans;
      else
        cout << " " << ans;
    }
    cout << endl;
  }
  return 0;
}
