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 int A[40005];
long long int _gcd(long long int a, long long int b) {
  return b == 0 ? a : _gcd(b, a % b);
}
long long int _lcm(long long int a, long long int b) {
  return a * b / _gcd(a, b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  long long int i, j, k, x, y, z, a, b, l, r, q, t;
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    long long int n = _lcm(a, b);
    for (i = 1; i <= n; ++i) {
      x = (i % a) % b;
      y = (i % b) % a;
      A[i] = x != y;
      A[i] += A[i - 1];
    }
    while (q--) {
      cin >> l >> r;
      x = r % n;
      y = r / n;
      a = y * A[n] + A[x];
      x = (l - 1) % n;
      y = (l - 1) / n;
      b = y * A[n] + A[x];
      cout << a - b << " ";
    }
    cout << endl;
  }
  return 0;
}
