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 int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    long long int q;
    cin >> q;
    long long int G = gcd(a, b);
    for (int i = 0; i < q; i++) {
      long long int l, r, ans = 0;
      cin >> l >> r;
      if (G != 1) {
        cout << 0 << " ";
        continue;
      }
      long long int k = (r - l + 1) / (a * b);
      if (k > 0) {
        ans += ((k * (a * b)) - k * max(a, b));
      }
      k = (r - l + 1) % (a * b);
      k = r - k + 1;
      for (long long int j = k; j <= r; j++) {
        if ((j % a) % b != (j % b) % a) ans++;
      }
      cout << ans << " ";
    }
    cout << endl;
  }
}
