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 mod = 1000000007;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  int ts;
  cin >> ts;
  for (long long(ti) = 0; (ti) < (ts); (ti)++) {
    long long a, b, q;
    cin >> a >> b >> q;
    vector<long long> ans(q);
    if (a > b) swap(a, b);
    long long num = lcm(a, b);
    int cnt = 0;
    for (long long(i) = 0; (i) < (num); (i)++) {
      int left = (i % a) % b;
      int right = (i % b) % a;
      if (left != right) cnt++;
    }
    for (long long(i) = 0; (i) < (q); (i)++) {
      long long l, r;
      cin >> l >> r;
      int sum = r / num * cnt + max(0LL, r % num - (num - cnt) + 1);
      sum -= l / num * cnt + max(0LL, l % num - (num - cnt));
      ans[i] = sum;
    }
    for (long long(i) = 0; (i) < (q); (i)++)
      printf("%d%c", ans[i], (i == q - 1 ? '\n' : ' '));
  }
}
