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>
#pragma GCC optimize("Ofast")
#pragma loop_opt(on)
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using ld = long double;
mt19937 mtrd(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 1000000007;
const int mod2 = 998244353;
const ld PI = acos(-1);
long long qpow(long long x, long long powcnt, long long tomod) {
  long long res = 1;
  for (; powcnt; powcnt >>= 1, x = (x * x) % tomod)
    if (1 & powcnt) res = (res * x) % tomod;
  return (res % tomod);
}
long long inv(long long x) { return qpow(x, mod - 2, mod); }
long long a, b, q;
tuple<long long, long long, long long> exgcd(long long x, long long y) {
  if (y == 0) return make_tuple(1, 0, x);
  long long X, Y, G;
  tie(X, Y, G) = exgcd(y, x % y);
  return make_tuple(Y, X - (x / y) * Y, G);
}
void solve() {
  cin >> a >> b >> q;
  if (a > b) swap(a, b);
  vector<long long> v;
  v.push_back(0);
  for (long long i = 1; i < a * b; i++)
    v.push_back(v.back() + ((i % a) != ((i % b) % a)));
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long ans = 0;
    long long x = l % (a * b);
    if (l / (a * b) == r / (a * b)) {
      long long y = r % (a * b);
      ans = v[y] - (x ? v[x - 1] : 0);
    } else {
      if (x) ans += v.back() - v[x - 1];
      l = (l + a * b - 1) / (a * b);
      l *= a * b;
      ans += (r - l) / (a * b) * v.back();
      ans += v[r % (a * b)];
    }
    cout << ans << ' ';
  }
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
}
