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;
int a, b, q;
vector<long long int> pre;
long long int solve(long long int n) {
  return (n / a * b) * pre[a * b - 1] + pre[n % (a * b)];
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    pre.clear();
    cin >> a >> b >> q;
    long long int sum = 0, ans;
    for (int i = 0; i < a * b; i++) {
      if (i % a % b != i % b % a) sum++;
      pre.push_back(sum);
    }
    while (q--) {
      long long int r, l;
      cin >> l >> r;
      ans = abs(((l - 1) / (a * b)) * pre[a * b - 1] + pre[(l - 1) % (a * b)] -
                ((r / (a * b)) * pre[a * b - 1] + pre[r % (a * b)]));
      cout << ans << "\n";
    }
  }
}
