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 pref(long long x, vector<long long> &d, long long a, long long b) {
  long long res = ((long long)d.size()) * (x / (a * b));
  res += upper_bound(d.begin(), d.end(), x % (a * b)) - d.begin();
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    int q;
    cin >> a >> b >> q;
    vector<long long> d;
    for (long long i = 0; i < a * b; ++i) {
      if ((i % a) % b != (i % b) % a) d.push_back(i);
    }
    for (int w = 0; w < q; ++w) {
      long long l, r;
      cin >> l >> r;
      cout << pref(r, d, a, b) - pref(l - 1, d, a, b) << ' ';
    }
    cout << '\n';
  }
}
