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;
int main() {
  long long int n, i, j, a, b, x, p, q, l, m, t;
  cin >> t;
  for (i = 0; i < t; i++) {
    cin >> n, x, p, q;
    a = min(p, q);
    b = max(p, q);
    l = n - b;
    m = a - 1;
    if ((l == 0) && (m == 0)) {
      cout << n - 1 << endl;
    } else if ((l < x) && (m < x)) {
      if (x - l > m) {
        cout << (b + l - 1) << endl;
      } else if (x - l < m) {
        cout << (b - a + x) << endl;
      }
    }
    if ((l > x) || (m > x)) {
      cout << b - a + x;
    }
  }
  return 0;
}
