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 func(long long x, long long j) {
  long long ans = 0;
  ans += x / 400 - 1;
  if (j <= x % 400) ans++;
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long l[q], r[q], i, j;
    for (i = 0; i < q; i++) {
      long long ans = 0;
      cin >> l[i] >> r[i];
      long long k = max(a, b);
      if (r[i] <= 400) {
        for (j = 1; j <= 400; j++) {
          if ((j % a) % b != (j % b) % a) {
            if (j >= l[i] && j <= r[i]) ans++;
          }
        }
      } else if (l[i] - 1 <= 400) {
        for (j = 0; j < 400; j++) {
          if (((j + 400) % a) % b != ((j + 400) % b) % a) ans += func(r[i], j);
          if (j == 0 && l[i] - 1 == 400) ans--;
        }
        for (j = 1; j < 400; j++) {
          if ((j % a) % b != (j % b) % a) {
            if (j >= l[i] && j <= r[i]) ans++;
          }
        }
      } else {
        for (j = 0; j < 400; j++) {
          if (((j + 400) % a) % b != ((j + 400) % b) % a) {
            ans += -func(l[i] - 1, j) + func(r[i], j);
          }
        }
      }
      cout << ans << " ";
    }
    cout << endl;
  }
}
