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>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long mxn = 200 * 200 + 10;
long long dp[mxn];
long long len;
void precompute(long long a, long long b) {
  len = a * b;
  for (long long i = 1; i <= len; i++) {
    dp[i] = dp[i - 1];
    if ((i % a) % b != (i % b) % a) {
      dp[i]++;
    }
  }
}
long long query(long long x) {
  long long rep = x / len;
  long long rem = x % len;
  return dp[len] * rep + dp[rem];
}
long long query(long long l, long long r) { return query(r) - query(l - 1); }
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    precompute(a, b);
    while (q--) {
      long long l, r;
      cin >> l >> r;
      cout << query(l, r) << " ";
    }
    cout << "\n";
  }
  return 0;
}