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;
const int INF = 1001001001;
const long long int LINF = 100100100100100100;
const double EPS = 1e-10;
const long double PI = acos(-1.0L);
__attribute__((constructor)) void initial() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
signed main() {
  int q;
  cin >> q;
  while (q--) {
    long long int a, b, q;
    cin >> a >> b >> q;
    long long int n = a * b;
    long long int f[n + 1];
    f[0] = 0LL;
    for (int i = 1; i < n; i++) {
      f[i] = f[i - 1];
      if (i % a % b != i % b % a) f[i]++;
    }
    while (q--) {
      long long int l, r;
      cin >> l >> r;
      cout << f[n - 1] * (r / n) + f[r % n] - f[n - 1] * ((l - 1) / n) -
                  f[(l - 1) % n]
           << ' ';
    }
    cout << endl;
  }
  return 0;
}
