Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int cnt[40005];
int t, a, b, q;
long long calc(long long x) {
  return x / (a * b) * cnt[a * b - 1] + cnt[x % (a * b)];
}
int main() {
  long long l, r;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d%d", &a, &b, &q);
    for (int i = 1; i < a * b; i++) {
      cnt[i] = cnt[i - 1];
      if (i % a % b != i % b % a) cnt[i]++;
    }
    while (q--) {
      scanf("%lld%lld", &l, &r);
      if (q)
        printf("%lld ", calc(r) - calc(l - 1));
      else
        printf("%lld\n", calc(r) - calc(l - 1));
    }
  }
  return 0;
}