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;
string char_to_str(char c) {
  string tem(1, c);
  return tem;
}
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  }
  return b;
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  }
  return b;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long lcm1 = lcm(a, b);
    long long max1 = max(a, b);
    long long last_num_tot = max1 - 1;
    for (int i = 0; i < q; ++i) {
      long long ans = 0;
      long long start, end;
      cin >> start >> end;
      long long q1 = start / lcm1;
      long long q2 = end / lcm1;
      if (q1 != q2) {
        long long bet = q2 - q1 - 1;
        ans = ans + bet * (last_num_tot + 1);
        long long last_one = last_num_tot + (q2 * lcm1);
        if (last_one <= end) {
          ans = ans + last_num_tot + 1;
        } else {
          long long l_first_one = q2 * lcm1;
          ans = ans + (end - l_first_one) + 1;
        }
        long long last_first_one = last_num_tot + (q1 * lcm1);
        if (last_first_one >= start) {
          ans = ans + (last_first_one - start) + 1;
        }
        long long last_last_first_one = q2 * lcm1;
        if (last_first_one >= last_last_first_one) {
          ans = ans - (last_first_one - last_last_first_one) - 1;
        }
      } else {
        long long q = q1;
        long long start_one = lcm1 * q;
        long long last_one = last_num_tot + (lcm1 * q);
        if ((start_one > end) || (last_one < start)) {
          ans = 0;
        } else {
          if ((start_one >= start) && (last_one <= end)) {
            ans = last_one - start_one + 1;
          } else if ((start_one < start) && (last_one > end)) {
            ans = end - start + 1;
          } else {
            if (last_one >= start) {
              ans = ans + (last_one - start) + 1;
            } else {
              ans = ans + (end - start_one) + 1;
            }
          }
        }
      }
      long long fin = end - start - ans + 1;
      if (fin <= 0) {
        cout << 0 << " ";
      } else {
        cout << end - start - ans + 1 << " ";
      }
    }
    cout << endl;
  }
  return 0;
}