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;
long long compute(vector<int> pre, long long l, long long a, long long b) {
  long long ans = (l / (a * b)) * ((long long)pre[a * b - 1]);
  long long i = (l) % (a * b);
  ans += (long long)pre[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    vector<int> pre(a * b, 0);
    if (a > b) swap(a, b);
    for (int i = 0; i < a * b; i++) {
      if ((i % b) % a != i % a) pre[i] = 1;
    }
    for (int i = 1; i <= a * b - 1; i++) pre[i] += pre[i - 1];
    while (q--) {
      long long l, r;
      cin >> l >> r;
      if (a == 1 && b == 1) {
        cout << 0 << " ";
        continue;
      }
      cout << compute(pre, r, a, b) - compute(pre, l - 1, a, b) << " ";
    }
    cout << "\n";
  }
}
