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>
using namespace std;
pair<long long, long long> add_pair(pair<long long, long long> a,
                                    pair<long long, long long> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
long long power(long long a, long long b) {
  long long res = 1;
  a = a % ((long long)1e9 + 7);
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % ((long long)1e9 + 7);
      b--;
    }
    a = (a * a) % ((long long)1e9 + 7);
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, ((long long)1e9 + 7) - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long mul(long long a, long long b) {
  return (a * (b % ((long long)1e9 + 7))) % ((long long)1e9 + 7);
}
long long sub(long long a, long long b) {
  return ((a - b) % ((long long)1e9 + 7) + ((long long)1e9 + 7)) %
         ((long long)1e9 + 7);
}
long long add(long long a, long long b) {
  return ((a + b) % ((long long)1e9 + 7) + ((long long)1e9 + 7)) %
         ((long long)1e9 + 7);
}
long long ctr = 0;
void solve() {
  long long a, b, q, l, r, x, g;
  cin >> a >> b;
  if (a > b) swap(a, b);
  if (b % a == 0) {
    cin >> q;
    while (q--) {
      ctr++;
      cin >> l >> r;
      if (ctr == 48011) cout << l << " ";
      cout << 0 << ' ';
    }
    cout << '\n';
    return;
  }
  g = gcd(a, b);
  a /= g;
  x = a * b;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    ctr++;
    if (ctr == 48011) cout << l << " ";
    long long sum = 0;
    if (r < b) {
      cout << 0 << ' ';
      continue;
    } else if (l < b)
      l = b;
    sum = ((r / x) - (l / x)) * b;
    sum = (r - l + 1) - sum;
    cout << sum << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
