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,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long maxn = 2e5 + 20;
const long long inf = (long long)2e9;
const long double pi = asin(1) * 2;
const long double eps = 1e-8;
const long long mod = (long long)1e9 + 7;
const long long ns = 97;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  long long a, b, q;
  cin >> a >> b >> q;
  if (a > b) swap(a, b);
  while (q--) {
    long long l, r;
    cin >> l >> r;
    if (b % a == 0 || r <= b) {
      cout << 0 << ' ';
    } else {
      l = max(l, b);
      long long ans = r - l + 1;
      long long mn = l / b, mx = r / b;
      long long A = 0, B = 0;
      for (long long i = mn; i < mx + 1; ++i) {
        if (b * i % a == 0) {
          A = i;
          break;
        }
      }
      for (long long i = mx; i >= mn; --i) {
        if (b * i % a == 0) {
          B = i;
          break;
        }
      }
      long long k = (B - A) / a + 1;
      ;
      ;
      ;
      if (B > 0) ans -= k * b;
      --l;
      while (l % b % a == l % a % b) {
        ++ans;
        --l;
      }
      ++r;
      while (r % b % a == r % a % b) {
        ++ans;
        ++r;
      }
      cout << ans << ' ';
    }
  }
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long q = 1;
  cin >> q;
  while (q--) solve();
}
