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;
int main(int argc, char const *argv[]) {
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q, l, r;
    cin >> a >> b >> q;
    long long i;
    if (a > b) swap(a, b);
    for (i = 0; i < q; i++) {
      cin >> l >> r;
      long long c = a * b;
      long long m = r - l + 1;
      long long neg = 0;
      while ((l % a) % b == (l % b) % a) {
        if (l == r) break;
        neg++;
        l++;
      }
      while ((r % a) % b == (r % b) % a) {
        neg++;
        r--;
        if (r <= l) break;
      }
      long long d = (l / c);
      long long e = r / c;
      neg += (e - d) * b;
      printf("%lld", m - neg);
      if (i < q - 1)
        printf(" ");
      else
        printf("\n");
    }
  }
  return 0;
}
