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;
int mod(string s, int a) {
  int ans = 0;
  for (int i = 0; i < s.length(); i++) ans = (ans * 10 + (int)s[i] - '0') % a;
  return ans;
}
long long min(long long x, long long y) {
  if (x < y) return x;
  return y;
}
long long max(long long x, long long y) {
  if (x > y) return x;
  return y;
}
long long power(long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
long long hcf(long long a, long long b) {
  if (b == 0) return a;
  return hcf(b, a % b);
}
pair<long long, long long> f(long long n, long long a, long long b) {
  long long x, c = 0, lcm = (a * b) / hcf(a, b), f, ans = n;
  if (n < b) return {0, 0};
  ans -= b - 1;
  x = n / lcm;
  if (n < x * lcm + b) {
    c += (x - 1) * (b - 1);
    ans -= c;
    c = n - (x * lcm);
    ans -= c;
  } else {
    ans -= (x) * (b - 1);
    f = 1;
  }
  return {ans, f};
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long TT = 1;
  cin >> TT;
  while (TT--) {
    long long a, b, q;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    while (q--) {
      long long l, r, i, flag, c;
      cin >> l >> r;
      if (b % a == 0)
        cout << "0 ";
      else {
        pair<long long, long long> p = f(l, a, b), q = f(r, a, b);
        c = q.first - p.first;
        cout << c << " ";
      }
    }
    cout << "\n";
  }
}
