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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("no-stack-protector")
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using str = string;
using pll = pair<ll, ll>;
const ll INF = 0x7fffffffffffffffLL;
const ll BASE = 1000000007LL;
template <class T>
T MAX() {
  return std::numeric_limits<T>::max();
}
template <class T>
T MIN() {
  return std::numeric_limits<T>::min();
}
template <class T>
void chMax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chMin(T &a, T b) {
  if (b < a) a = b;
}
inline ll divUp(ll a, ll b) { return (a + b - 1) / b; }
void solve() {
  ll a, b, q;
  cin >> a >> b >> q;
  vector<ll> preSum(a * b);
  for (ll i = 0; i < a * b; i++) {
    if (i % a % b != i % b % a) preSum[i]++;
  }
  for (ll i = 0; i < preSum.size() - 1; i++) {
    preSum[i + 1] += preSum[i];
  }
  auto ask = [&](ll v) {
    return (v / preSum.size() * preSum.back()) + preSum[v % preSum.size()];
  };
  while (q--) {
    ll l, r;
    cin >> l >> r;
    cout << ask(r) - ask(l - 1) << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << setprecision(10) << fixed;
  ll t;
  cin >> t;
  while (t--) solve();
  return 0;
}