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 gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long c = min(a, b);
    long long d = max(a, b);
    while (q--) {
      long long l, r;
      cin >> l >> r;
      long long lcm = a * b / gcd(a, b);
      long long div1 = l / lcm, div2 = r / lcm;
      long long di = div2 - div1;
      long long ans = r - l + 1;
      if (di >= 1) ans = ans - (d * (di - 1));
      long long x = 0;
      if (di >= (long long)1) x = min((r % lcm) + 1, d);
      long long y = 0;
      if (div1 >= (long long)1) {
        if ((l % lcm) + 1 <= d) {
          y = d - (l % lcm) + 1;
        } else
          y = 0;
      }
      for (long long i = l; i <= min(d, r); i++) {
        if ((i % a) % b == (i % b) % a) ans--;
      }
      ans = ans - x - y;
      cout << ans << " ";
    }
    cout << endl;
  }
}
