Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 998244353;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 998244353;
      b--;
    }
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 998244353 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  cin >> T;
  while (T--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long d = (a * b) / gcd(a, b);
    vector<long long> v(d, 0);
    for (long long i = 0; i < d; i++) {
      if ((i % a) % b != (i % b) % a) v[i]++;
      if (i > 0) v[i] += v[i - 1];
    }
    while (q--) {
      long long l, r;
      cin >> l >> r;
      long long ans = 0;
      ans -= v[d - 1] * ((l - 1) / d) + v[(l - 1) % d];
      ans += v[d - 1] * ((r) / d) + v[(r) % d];
      cout << ans << " ";
    }
    cout << "\n";
  }
  return 0;
}