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 gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t, z = 1;
  cin >> t;
  while (t--) {
    long long int a, b, q;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    while (q--) {
      long long int l, r;
      cin >> l >> r;
      long long int lcm = (a * b) / (gcd(a, b));
      if (r <= 100000) {
        long long int ans = 0;
        for (long long int i = l; i <= r; i++) {
          if ((i % a) % b != (i % b) % a) ans++;
        }
        cout << ans << " ";
      } else {
        long long int x = l / lcm, y = r / lcm;
        if (l % lcm != 0) x++;
        if ((x * lcm) > r || (y * lcm) < l) {
          cout << (r - l + 1) << " ";
          continue;
        }
        long long int ans = 0;
        if (l % lcm != 0) {
          for (long long int i = l; i <= min((l + a), (x * lcm) - 1); i++) {
            if ((i % a) % b == (i % b) % a) ans++;
          }
        }
        long long int g = 0;
        long long int h = y * lcm;
        g = (min(r, h + b - 1) - h + 1);
        cout << (r - l + 1 - ans - (y - x) * b - g) << " ";
      }
    }
    cout << "\n";
  }
}
