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;
void solve() {
  long long a, b, q;
  cin >> a >> b >> q;
  if (a == 1 || b == 1) {
    for (long long i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      cout << 0 << ' ';
    }
    cout << "\n";
    return;
  }
  vector<long long> ans;
  for (long long i = 1; i <= a * b; i++) {
    ans.push_back(((i % a) % b) != ((i % b) % a));
  }
  vector<long long> pref(a * b + 1);
  for (long long i = 0; i < a * b; i++) {
    pref[i + 1] = pref[i] + ans[i];
  }
  for (long long i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    long long now_ans = 0;
    if ((r - l + 1) > a * b) {
      long long new_r = r - (r % (a * b));
      long long new_l = l + ((a * b - (l % (a * b))) % (a * b));
      now_ans += (new_r - new_l + 1) / (a * b) * pref[a * b];
      l %= a * b;
      now_ans += pref[a * b] - pref[l - 1];
      r %= a * b;
      now_ans += pref[r];
    } else {
      now_ans += pref[r % (a * b + 1)] - pref[(l - 1) % (a * b + 1)];
    }
    cout << now_ans << ' ';
  }
  cout << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
