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 div_up(long long x, long long y) {
  if (x == 0) return 0;
  return (x - 1) / y + 1;
}
bool is_prime(long long a) {
  for (long long i = 2; i <= sqrt(a); i++)
    if (a % i == 0) return false;
  return true;
}
bool borders(int x, int y, int n, int m) {
  return x >= 0 && y >= 0 && x < n && y < m;
}
long long aba(vector<int> f, long long pos, long long n) {
  return f[n - 1] * (pos / n) + f[pos % n];
}
void solve() {
  long long a, b, q, l, r;
  cin >> a >> b >> q;
  vector<int> f(a * b);
  for (int i = 1; i < a * b; i++) {
    f[i] = f[i - 1];
    if ((i % a) % b != (i % b) % a) f[i]++;
  }
  vector<long long> ans;
  while (q--) {
    cin >> l >> r;
    cout << (aba(f, r, a * b) - aba(f, l - 1, a * b)) << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
