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;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, x, arr[40002], i, c = 0, mul;
    cin >> a >> b >> x;
    mul = a * b;
    for (i = 1; i <= a * b; i++) {
      if ((i % a) % b != (i % b) % a) c++;
      arr[i] = c;
    }
    arr[0] = 0;
    while (x--) {
      c = 0;
      long long p1, p2, total, per, rem, p, sum, f1, f2;
      cin >> p1 >> p2;
      f1 = (p1 - 1) / mul + 1;
      f1 *= mul;
      f2 = p2 / mul;
      f2 *= mul;
      sum = (f2 - f1) / mul * arr[mul];
      if (f1 != p1) sum += arr[mul] - arr[p1 % mul - 1];
      if (f2 != p2) sum += arr[p2 - f2];
      cout << sum << " ";
    }
    cout << endl;
  }
}
