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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long a1, b1, q;
    cin >> a1 >> b1 >> q;
    long long a = min(a1, b1), b = max(a1, b1);
    long long h = (a * b) / gcd(b, a);
    for (long long j = 0; j < q; j++) {
      long long l, r, x, y, k = 0;
      cin >> l >> r;
      x = l / h;
      y = r / h;
      if (x != y) {
        k = (y - x - 1) * b;
        k += (x * h + b - 1 >= l ? x * h + b - l : 0);
        k += (y * h + b - 1 <= r ? b : r - y * h + 1);
        cout << r - l - k + 1 << " ";
      } else {
        if (x * h + b - 1 < l || x * h > r) {
          k = 0;
          cout << r - l - k + 1 << " ";
        } else {
          k += (min(r, x * h + b - 1) - max(l, x * h) + 1);
          cout << r - l - k + 1 << " ";
        }
      }
    }
    cout << endl;
  }
  return 0;
}
