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 (a == 0) return b;
  return gcd(b % a, a);
}
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b, q, l, r;
    cin >> a >> b;
    cin >> q;
    long long int mul = a * b;
    long long int xx = lcm(a, b);
    while (q--) {
      cin >> l >> r;
      long long int total;
      long long int m1 = max(a, b);
      long long int m2 = min(a, b);
      if (r < m1) {
        cout << 0 << " ";
      } else {
        if (l < m1) {
          l = m1;
        }
        total = (r - l + 1);
        long long int k = l % xx;
        long long int f = 0;
        if (k < m1) {
          l += (m1 - k);
          total -= (m1 - k);
          if (r < l) {
            f = 1;
            total += (l - (r + 1));
          }
        }
        if (!f) {
          k = r % xx;
          if (k < m1) {
            r -= (k + 1);
            total -= (k + 1);
            if (r < l) total += (l - (r + 1));
          }
        }
        if (l < r) {
          long long int x1 = r / xx;
          long long int x2 = (l - 1) / xx;
          long long int t2 = x1 - x2;
          total -= t2 * (m1);
        }
        cout << total << " ";
      }
    }
    cout << "\n";
  }
  return 0;
}
