Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int t;
long long int a, b, q;
long long int l, r;
long long int ans[510];
long long int gcd(long long int n, long long int m) {
  if (m == 0) return n;
  if (n < m) return gcd(m, n);
  return gcd(m, n % m);
}
long long int lcm(long long int n, long long int m) {
  return n * m / gcd(n, m);
}
long long int count(long long int n) {
  long long int x = n / lcm(a, b);
  long long int y = n % lcm(a, b);
  x *= b;
  if (y >= b) y = b - 1;
  return x + y;
}
int main() {
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    if (a > b) {
      int tmp = a;
      a = b;
      b = tmp;
    }
    long long int m = lcm(a, b);
    for (int i = 1; i <= q; i++) {
      cin >> l >> r;
      long long int s = count(r) - count(l - 1);
      ans[i] = (r - l + 1) - s;
    }
    for (int i = 1; i <= q; i++) cout << ans[i] << " ";
    cout << endl;
  }
}