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;
int main() {
  unsigned long long int t, a, b, q, l, r, cnt, c;
  cin >> t;
  while (t--) {
    vector<unsigned long long int> cont;
    cin >> a >> b >> q;
    int arr[40001] = {0};
    c = a * b;
    for (int i = 0; i < c; i++) {
      if ((i % a) % b != (i % b) % a) arr[i] = 1;
    }
    unsigned long long int sum[40002];
    sum[0] = 0;
    for (int i = 1; i < c; i++) {
      sum[i] = sum[i - 1] + arr[i];
    }
    while (q--) {
      cin >> l >> r;
      cnt = (r / c) * sum[a * b - 1];
      cnt += sum[r % c];
      cnt = cnt - ((l / c) * sum[c - 1]);
      if (l % c == 0)
        cnt = cnt - sum[l % c];
      else
        cnt -= sum[l % c - 1];
      cont.push_back(cnt);
    }
    for (auto it : cont) cout << it << " ";
    cout << endl;
  }
}