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 t;
  scanf("%d", &t);
  while (t--) {
    long long int a, b, q, l, r, sum = 0, g, k;
    scanf("%lld%lld%lld", &a, &b, &q);
    for (int i = 1; i <= q; i++) {
      scanf("%lld%lld", &l, &r);
      g = (a * b);
      sum = r - max(a, b) + 1;
      if (l > max(a, b)) sum = sum - (l - max(a, b));
      if (sum <= 0)
        printf("0 ");
      else {
        if (l % g == 0) {
          if (r - l + 1 >= max(a, b)) {
            sum = sum - max(a, b);
            l = l + max(a, b);
            sum = sum - ((r - l + 1) / g) * max(a, b);
          } else
            sum = sum - (r - l);
        } else {
          if (l >= ((l / g) * g + max(a, b)) && (r >= g)) {
            if (r >= ((l / g + 1) * g)) {
              k = r - (l / g + 1) * g + 1;
              if (k >= max(a, b)) {
                sum = sum - max(a, b);
                l = (l / g + 1) * g + max(a, b);
                sum = sum - ((r - l + 1) / g) * max(a, b);
              } else {
                sum = sum - k;
              }
            }
          } else {
            if (r - l + 1 >= max(a, b) && (r >= g)) {
              cout << r << " " << l << " " << max(a, b) << endl;
              sum = sum - max(a, b) + (l - (l / g) * g);
              l = l + max(a, b);
              sum = sum - ((r - l + 1) / g) * max(a, b);
            } else if (r >= g)
              sum = sum - (r - l + 1) + (l - (l / g) * g);
          }
        }
        if (sum <= 0)
          printf("0 ");
        else
          printf("%lld ", sum);
      }
      printf("\n");
    }
  }
  return 0;
}
