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;
long long fact(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return fact(b, a % b);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int i, j, k, n, a, b, q, w;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    w = (a * b) / fact(a, b);
    while (q--) {
      long long int l, r, k = 0, y, x;
      cin >> l >> r;
      if (l > r) swap(l, r);
      if (r <= w) {
        if (r > b)
          cout << r - b + 1 << " ";
        else
          cout << "0"
               << " ";
      } else {
        j = (r / w - l / w) * (w - b);
        x = r % w;
        y = l % w;
        if (x >= b) j = j + x - b + 1;
        if (y >= b) j = j - (y - b);
        cout << abs(j) << " ";
      }
    }
    cout << endl;
  }
  return 0;
}
