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;
const int N = 1e2 + 5;
const long long mod = 1e9 + 7;
long long powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y % 2 == 1) z = (z * x);
    y /= 2;
    x = (x * x);
  }
  return z;
}
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 (max(a, b) / gcd(a, b)) * min(a, b);
}
long long calc(long long x, long long a, long long b) {
  long long l = (a * b) / gcd(a, b);
  if (a < b) swap(a, b);
  long long ans = (x / l - 1) * a + min(x % l + 1, a);
  return ans;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    while (q--) {
      long long l, r, m;
      cin >> l >> r;
      cout << r - l + 1 - calc(r, a, b) + calc(l - 1, a, b) << " ";
    }
    cout << endl;
  }
  return 0;
}
