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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const unsigned long MOD = 1e9 + 7;
int t, a, b, q, p, m, pmi, i;
unsigned long l, r, pm, res;
void solve() {
  cin >> a >> b >> q;
  if (a < b)
    p = b, m = a;
  else
    p = a, m = b;
  while (q--) {
    cin >> l >> r;
    if (a == b)
      cout << "0 ";
    else {
      if (p > r)
        cout << "0 ";
      else {
        if (p > l) l = p;
        res = (r - l + 1);
        pm = p * m;
        pm = pm * floor(r / pm);
        while (pm >= l) {
          res -= 1 + min(r, pm + p - 1) - pm;
          pm -= pm;
        }
        cout << res << " ";
      }
    }
  }
  cout << "\n";
}
int main() {
  cin >> t;
  while (t--) solve();
  return 0;
}
