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;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long a, b, c;
    scanf("%lld%lld%lld", &a, &b, &c);
    long long maxi = max(a, b);
    for (long long i = 0; i < c; i++) {
      long long l, r;
      long long sum = 0;
      scanf("%lld%lld", &l, &r);
      sum = r - l + 1;
      long long diffa = l / lcm(a, b);
      long long diffb = r / lcm(a, b);
      long long cnt = 0;
      if (diffa * lcm(a, b) + maxi < l) {
        cnt++;
      }
      if (diffb * lcm(a, b) + maxi > r) {
        cnt++;
      }
      sum -= max(diffb - diffa - cnt, (long long)0) * maxi;
      for (long long k = 0; k < maxi; k++) {
        if (diffa * lcm(a, b) + k >= l && diffa * lcm(a, b) + k <= r) {
          sum--;
        }
      }
      if (diffa < diffb && cnt == 1) {
        for (long long k = 0; k < maxi; k++) {
          if (diffb * lcm(a, b) + k >= l && diffb * lcm(a, b) + k <= r) {
            sum--;
          }
        }
      }
      printf("%lld ", sum);
    }
    printf("\n");
  }
  return 0;
}
