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;
int gcd(int a, int b) {
  while (a != b) {
    if (a > b)
      a -= (a / b) * b;
    else
      b -= (b / a) * a;
    if (!a) return b;
    if (!b) return a;
  }
  return a;
}
void solve() {
  cin >> a >> b >> q;
  int ta = a;
  a /= gcd(a, b);
  for (int i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    int low = max(ta, b);
    long long ans = 0;
    if (l == r) cout << (l >= low) << " ";
    if (r < (l / (a * b) + 1) * a * b)
      ans += max(r % (a * b) - low + 1, 0ll) - max(l % (a * b) - low, 0ll);
    else
      ans += a * b - low - max(l % (a * b) - low, 0ll);
    l = (l / (a * b) + 1) * a * b;
    if (l < r) ans += max(r % (a * b) - low + 1, 0ll);
    r = r / (a * b) * a * b;
    cout << ans << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}
