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 n = 1000000000000;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    int x, y, q;
    cin >> x >> y >> q;
    vector<long long> v;
    long long maxi = 0, mini = 1;
    v.push_back(0);
    while (q--) {
      long long l, r;
      cin >> l >> r;
      if (r >= maxi) {
        for (long long i = max(mini, maxi + 1); i < r + 1; i++) {
          if (i <= x || i < y) {
            v.push_back(0);
            continue;
          }
          if ((i % x) % y != (i % y) % x)
            v.push_back(v[i - 1] + 1);
          else
            v.push_back(v[i - 1]);
        }
        maxi = max(r, maxi);
      }
      cout << v[r] - v[l - 1] << " ";
    }
    cout << endl;
  }
}
