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;
char _;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  ios_base::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;
    long long temp = a;
    a = max(a, b);
    b = min(temp, b);
    for (int i = 0; i < q; i++) {
      long long l, r, count = 0;
      cin >> l >> r;
      long long diff = r - l + 1;
      while (l <= r && l < a && (l % a != 0 || l % b != 0)) {
        if ((l % a) % b == (l % b) % a) {
          count++;
        }
        l++;
      }
      long long lcm = ::lcm(a, b);
      long long bigMul = (r - l + 1) / lcm;
      if (bigMul == 0) {
        while (l <= r && (l % a) % b == (l % b) % a) {
          l++;
          count++;
        }
        cout << diff - count << " ";
        continue;
      }
      count += bigMul * a;
      l += lcm * bigMul;
      while (l <= r) {
        if ((l % a) % b == (l % b) % a) {
          count++;
        }
        l++;
      }
      cout << diff - count << " ";
    }
    cout << "\n";
  }
  return 0;
}
