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;
int t, a, b, q, ans = 0, nww;
long long l, r, razem, odejmij;
int NWD(int x, int y) {
  if (x % y == 0) return y;
  return NWD(y, x % y);
}
int NWW(int x, int y) { return (x * y) / NWD(x, y); }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    if (b < a) swap(a, b);
    nww = NWW(a, b);
    for (int i = 0; i < q; i++) {
      cin >> l >> r;
      razem = (r / nww) * b;
      for (int j = 0; j < b; j++) {
        if ((((r / nww) * nww) + j) > r) razem--;
      }
      odejmij = (l / nww) * b;
      for (int j = 0; j < b; j++) {
        if ((((l / nww) * nww) + j) >= l) odejmij--;
      }
      cout << r - l + 1 - (razem - odejmij) << " ";
    }
    cout << endl;
  }
  return 0;
}
