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 abso(long x) {
  if (x < 0)
    return (-x);
  else
    return (x);
}
int main() {
  long x, y, a, b, c, d, t, n, i, j, k, a2, b2, q;
  long p;
  long long l, r;
  cin >> t;
  long res[40000];
  long sousres, sol;
  long flag, s, reste;
  long long taille;
  while (t) {
    t--;
    cin >> a >> b >> q;
    p = a * b;
    sousres = 0;
    for (i = 0; i < p; i++)
      if (((i % b) % a) != ((i % a) % b)) {
        res[i] = 1;
        sousres++;
      } else
        res[i] = 0;
    for (i = 0; i < q; i++) {
      cin >> l >> r;
      taille = r - l + 1;
      reste = (taille % p);
      sol = 0;
      for (j = 0; j < reste; j++) {
        sol += res[(p + r - j) % p];
      }
      sol += ((taille - reste) / p) * sousres;
      cout << sol << endl;
    }
  }
  return 0;
}
