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>
using namespace std;
const int N = 2e5 + 10;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long solve(long long x, long long d, long long num) {
  long long res = x / d * num, left = x / d * d;
  return res + min(num, x - left + 1);
}
int main() {
  long long T;
  scanf("%lld", &T);
  while (T--) {
    long long a, b, q;
    scanf("%lld %lld %lld", &a, &b, &q);
    if (a > b) swap(a, b);
    long long d = a * b / gcd(a, b);
    while (q--) {
      long long x, y;
      cin >> x >> y;
      if (a == b) {
        puts("0");
        continue;
      }
      long long x0 = x % d, y0 = y - x + x0;
      if (x0 == 0)
        printf("%lld ", y - x + 1 - solve(y0, d, b));
      else
        printf("%lld ", y - x + 1 - solve(y0, d, b) + solve(x0 - 1, d, b));
    }
    cout << endl;
  }
  return 0;
}