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 func(long long value, long long poka, long long x, long long y) {
  long long res = 0;
  if (value < y) {
    return value;
  }
  long long p = value / poka;
  if (p > 0) {
    res = ((p - 1) * y) + y - 1;
    if (value % poka < y) {
      res += value % poka + 1;
    } else if (value % poka > y) {
      res += y;
    }
  } else {
    res = y - 1;
  }
  return res;
}
long long GCD(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b) return GCD(a - b, b);
  return GCD(a, b - a);
}
long long LCM(long long a, long long b) {
  long long res = (a * b) / GCD(a, b);
  return res;
}
int main() {
  int t;
  cin >> t;
  for (int ii = 0; ii < t; ii++) {
    long long x, y, z;
    cin >> x >> y >> z;
    long long poka = LCM(x, y);
    for (int i = 0; i < z; i++) {
      long long a, b;
      cin >> a >> b;
      long long res = 0;
      long long ans_1 = func(a, poka, x, y);
      long long ans_2 = func(b, poka, x, y);
      res = b - a - (ans_2 - ans_1);
      if ((a % x) % y != (a % y) % y && b - a > 1) {
        res += 1;
      }
      cout << res << endl;
    }
  }
  return 0;
}
