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 a, b, q;
bool work[400005];
int pre[400005], pos[400005];
void solve() {
  cin >> a >> b >> q;
  unsigned long long tot = 0;
  for (int i = 0; i < a * b; i++) {
    work[i] = (i % a) % b != (i % b) % a;
    tot += work[i];
  }
  for (int i = 0; i < a * b; i++) pos[i + 1] = pos[i] + work[i];
  for (int i = a * b - 1; i > -1; i--) pre[i] = pre[i + 1] + work[i];
  for (int i = 0; i < a * b; i++) cout << pre[i] << " ";
  cout << endl;
  for (int i = 0; i < q; i++) {
    unsigned long long l, r;
    cin >> l >> r;
    unsigned long long ans = 0;
    if ((l / a / b + 1) * a * b >= r && l % (a * b)) {
      cout << pre[l % (a * b)] - (r % (a * b) ? pre[r % (a * b) + 1] : 0)
           << " ";
      continue;
    } else {
      ans += pre[l % (a * b)];
      l = (l / a / b + 1) * a * b;
    }
    if (l != r) {
      ans += pos[r % (a * b) + 1];
      r = r / a / b * a * b;
    }
    ans += tot * (r - l) / a / b;
    cout << ans << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}
