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;
using lli = long long int;
using vi = vector<int>;
using vll = vector<long long int>;
int pref[500001];
int main() {
  int T;
  cin >> T;
  while (T--) {
    int A, B, Q;
    cin >> A >> B >> Q;
    pref[0] = 0;
    for (int i = 1; i < (A * B); i++) {
      pref[i] = pref[i - 1];
      if ((i % A) % B != (i % B) % A) {
        pref[i]++;
      }
    }
    while (Q--) {
      lli L, R;
      cin >> L >> R;
      int len = A * B;
      lli tillR = (R / len) * pref[len - 1] + pref[R % len];
      L--;
      lli tillL = (L / len) * pref[len - 1] + pref[L % len];
      cout << tillR - tillL << " ";
    }
    cout << "\n";
  }
  return 0;
}
