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) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, l, r, max, q;
    cin >> a >> b >> q;
    long long g = lcm(a, b);
    if (a > b)
      max = a;
    else
      max = b;
    while (q--) {
      cin >> l >> r;
      long long c = r - l + 1;
      long long p = (r / g) - (l / g);
      if (((l / g) * g + max - 1) >= l) c -= (((l / g) * g + max) - l);
      if (p >= 1) {
        c -= ((p - 1) * (max));
        long long tmp = r - ((r / g) * g);
        if (tmp >= max)
          c -= max;
        else
          c -= tmp + 1;
      }
      if (l < max) c -= (max - l);
      if (c < 0) c = 0;
      cout << c << " ";
    }
    cout << endl;
  }
}
