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 int gcd(long long int x, long long int y) {
  if (!y) return x;
  return gcd(y, x % y);
}
int main() {
  int t;
  cin >> t;
  int a, b;
  int q;
  int big, smol;
  while (t--) {
    cin >> a >> b >> q;
    if (a > b) {
      big = a;
      smol = b;
    } else {
      big = b;
      smol = a;
    }
    long long int l, r;
    long long int sub;
    long long int before_lcm;
    long long int last_lcm;
    long long int number_of_lcms;
    long long int lcm;
    long long int ans;
    while (q--) {
      cin >> l >> r;
      lcm = (big * smol) / (gcd(big, smol));
      before_lcm = lcm * (l / lcm);
      if (before_lcm == l)
        sub = 0;
      else {
        sub = (before_lcm + big - 1) - l + 1;
        if (sub < 0) sub = 0;
      }
      number_of_lcms = (r / lcm) - (l / lcm);
      sub = sub + (number_of_lcms)*big;
      last_lcm = lcm * (r / lcm);
      if (((last_lcm + big - 1) - r + 1) > 0)
        sub = sub - ((last_lcm + big - 1) - r);
      ans = r - l + 1 - sub;
      cout << ans << endl;
    }
  }
}
