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>
int main() {
  int t;
  int a, b, c, q;
  long long l, r;
  scanf("%d", &t);
  for (register int i = 0; i < t; ++i) {
    scanf("%d %d %d", &a, &b, &q);
    long long result = 0;
    c = a > b ? a : b;
    for (int j = 0; j < q; ++j) {
      scanf("%lld %lld", &l, &r);
      if (r < c || a == b) {
        printf("%d ", 0);
      } else if (l < c) {
        result = r - c + 1 - (r / (a * b)) * c;
        printf("%lld ", result);
      } else {
        if (r == l) {
          printf("%d ", 1);
        } else {
          result = r - l + 1 - ((r / (a * b) - l / (a * b)) * c);
          printf("%lld ", result);
        }
      }
    }
  }
}
