Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
pair<int, int> pii;
pair<long long, long long> pll;
vector<vector<int>> edge;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long a, b, q;
    cin >> a >> b >> q;
    int size = a * b;
    vector<int> pre(size);
    for (int i = 0; i < size; i++) {
      if (i != 0) pre[i] = pre[i - 1];
      if ((i % a) % b != (i % b) % a) pre[i]++;
    }
    while (q--) {
      long long l, r;
      cin >> l >> r;
      long long k1 = (l - 1) / size;
      k1 = 1LL * pre[(l - 1) % size] + k1 * 1LL * pre[size - 1];
      long long k2 = r / size;
      k2 = 1LL * pre[r % size] + k2 * 1LL * pre[size - 1];
      cout << k2 - k1 << " ";
    }
    cout << "\n";
  }
}