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 t;
int a, b, q;
long long l, r;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    int big = (a > b) ? a : b;
    int small = (a > b) ? b : a;
    vector<int> val1(a * b + 1);
    vector<int> val2(a * b + 1);
    for (int i = 1; i <= a * b; ++i) {
      val1[i] = i % small;
      val2[i] = (i % big) % small;
    }
    vector<long long> cnt(a * b + 1);
    for (int i = 1; i <= a * b; ++i) {
      cnt[i] = cnt[i - 1];
      if (val1[i] != val2[i]) {
        ++cnt[i];
      }
    }
    vector<long long> ans(q + 1);
    for (int i = 1; i <= q; ++i) {
      cin >> l >> r;
      long long q1 = l / (a * b);
      long long q2 = r / (a * b);
      if (q1 == q2) {
        ans[i] = cnt[r % (a * b)] - cnt[(l - 1) % (a * b)];
      } else {
        ans[i] = (cnt[cnt.size() - 1] - cnt[(l - 1) % (a * b)]) +
                 (q2 - q1 - 1) * cnt[cnt.size() - 1] + +cnt[r % (a * b)];
      }
    }
    for (int i = 1; i <= q; ++i) {
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}
