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 Lpow(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a) % 998244353;
    a = (a * a) % 998244353;
    b = b / 2;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int a, b, q;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    long long int lc = (a * b) / gcd(a, b);
    while (q--) {
      long long int l, r;
      cin >> l >> r;
      long long int first_lc = lc * (ceil((double)l / lc));
      long long int last_lc = lc * (r / lc);
      long long int num = (last_lc - first_lc) / lc;
      if (num == -1) {
        long long int cnt = 0;
        for (long long int i = l; i <= r; i++) {
          if (i % lc < b) cnt++;
        }
        cout << r - l + 1 - cnt << " ";
      } else {
        long long int cnt = num * b;
        for (long long int i = l; i < first_lc; i++) {
          if (i % lc < b) cnt++;
        }
        for (long long int i = last_lc; i <= r; i++) {
          if (i % lc < b) cnt++;
        }
        cout << r - l + 1 - cnt << " ";
      }
    }
    cout << '\n';
  }
  return 0;
}
