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 n, l, r, a, b, q;
long long sum(long long x) {
  long long k = a * b;
  if (x < b) return 0;
  if (x < a * b) return x - b + 1;
  long long t1 = (x / k - 1) * b;
  if (x % k < b)
    t1 += x % k + 1;
  else
    t1 += b;
  return x - b - t1 + 1;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    if (a > b) {
      long long k = b;
      b = a;
      a = k;
    }
    while (q--) {
      scanf("%lld%lld", &l, &r);
      if (b % a == 0 || (r < b))
        cout << "0 ";
      else {
        long long ans = sum(r) - sum(l - 1);
        cout << ans << ' ';
      }
    }
    cout << "\n";
  }
  return 0;
}
