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 n, a, b, q, c, d, sum, flag, go, da, e[100000];
  cin >> n;
  while (n--) {
    cin >> a >> b >> q;
    for (int i = 1; i <= q; i++) {
      go = a * b;
      cin >> c >> d;
      da = abs(c / go - d / go);
      if (min(c, d) % go == 0) da++;
      if (max(c, d) < go * (max(c, d) / go) + max(a, b)) {
        da = (da - 1) * max(a, b) + max(c, d) - go * (max(c, d) / go) + 1;
      } else
        da = da * max(a, b);
      if (max(c, d) > max(a, b) && min(c, d) < max(a, b)) {
        da = max(c, d) - max(a, b) - da + 1;
        e[i] = da;
      } else if (max(a, b) > max(c, d)) {
        e[i] = 0;
      } else
        e[i] = abs(c - d) - da + 1;
    }
    for (int i = 1; i <= q; i++) {
      cout << e[i] << " ";
    }
    cout << endl;
  }
}
