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;
using ll = long long;
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int a, b, q;
    cin >> a >> b >> q;
    vector<ll> x(lcm(a, b));
    for (int i = 1; i <= x.size(); i++) {
      if (i % a % b != i % b % a) x.at(i - 1)++;
    }
    for (int i = 1; i < x.size(); i++) x.at(i) += x.at(i - 1);
    while (q--) {
      ll l, r;
      cin >> l >> r;
      l--;
      ll low = 0, high = 0;
      low += l / lcm(a, b) * x.back();
      if (l % lcm(a, b) > 0) low += x.at(l % lcm(a, b) - 1);
      high += r / lcm(a, b) * x.back();
      if (r % lcm(a, b) > 0) high += x.at(r % lcm(a, b) - 1);
      cout << high - low;
      if (q > 0) cout << " ";
    }
    cout << '\n';
  }
}