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 ull = unsigned long long;
ull p[10000000], m;
void build(short a, short b) {
  p[0] = 0;
  m = a * b;
  for (int i = 1; i <= m; i++) {
    p[i] = p[i - 1];
    if ((i % a) % b != (i % b) % a) p[i]++;
  }
}
ull fun(ull s) {
  ull full = s / m;
  ull rem = s % m;
  return p[m] * full + p[rem];
}
ull fun(ull l, ull r) { return fun(r) - fun(l - 1); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    short a, b, q;
    cin >> a >> b >> q;
    build(a, b);
    ull l, r;
    while (q--) {
      cin >> l >> r;
      cout << fun(l, r) << endl;
    }
  }
  return 0;
}
