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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long nok, a, b, q;
long long nod(long long x, long long y) {
  while (x > 0 && y > 0) {
    if (x > y)
      x %= y;
    else
      y %= x;
  }
  return x + y;
}
long long f(long long x) {
  long long xx = x;
  long long ans = 0;
  if (x >= nok - 1) {
    x -= nok - 1;
    ans += max(a, b) - 1;
  } else {
    ans += min(x, max(a, b) - 1);
    return xx - ans;
  }
  ans += x / nok * max(a, b);
  x = x % nok;
  ans += min(x, max(a, b));
  return xx - ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long t;
  cin >> t;
  while (t > 0) {
    t--;
    cin >> a >> b >> q;
    nok = a * b / nod(a, b);
    for (long long i = 1; i <= q; i++) {
      long long l, r;
      cin >> l >> r;
      cout << f(r) - f(l - 1) << ' ';
    }
    cout << endl;
  }
}
