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 f[210 * 210];
int ab;
long long solve(long long n) {
  long long res = f[ab - 1] * (n / ab);
  int m = n % ab;
  res += f[m];
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    ab = a * b;
    memset(f, 0, sizeof f);
    for (int i = (1); i < (ab); ++i) f[i] += (i % a % b) != (i % b % a);
    for (int i = (1); i < (ab); ++i) f[i] += f[i - 1];
    int q;
    cin >> q;
    while (q--) {
      long long q1, q2;
      cin >> q1 >> q2;
      long long res = solve(q2) - solve(q1 - 1);
      cout << res << " ";
    }
    cout << endl;
  }
}
