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;
const int N = 201000;
const double eps = 1e-8;
const int inf = 1e9 + 7;
const long long mod = 1e9 + 7;
long long p[N];
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
void solve() {
  long long i, a, b, q, l, r, ans, g;
  cin >> a >> b >> q;
  if (a > b) swap(a, b);
  g = a / gcd(a, b) * b;
  for (i = 0; i < g; i++) p[i + 1] = p[i] + (i % a % b != i % b % a);
  while (q--) {
    cin >> l >> r;
    ans = (r - l + 1) / g * p[g];
    l %= g;
    r %= g;
    if (l <= r)
      ans += p[r + 1] - p[l];
    else
      ans += p[g] - p[l] + p[r + 1];
    cout << ans << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, n;
  cin >> n;
  while (n--) solve();
  return 0;
}
