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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    vector<long long> dp(a * b);
    long long count = 0;
    for (long long i = 0; i < a * b; ++i) {
      if (((i % a) % b) != ((i % b) % a)) count++;
      dp[i] = count;
    }
    for (long long i = 0; i < q; ++i) {
      long long l, r;
      cin >> l >> r;
      long long temp1 = (l + a * b - 1) / (a * b);
      long long temp2 = (r) / (a * b);
      long long start = temp1 * a * b - 1;
      long long hi = start % (a * b);
      long long lo = l % (a * b);
      long long g = 0, h = 0;
      if (hi >= lo) {
        if (lo >= 1)
          g = dp[hi] - dp[lo - 1];
        else
          g = dp[hi];
      }
      long long end = temp2 * a * b;
      hi = r % (a * b);
      lo = end % (a * b);
      if (hi >= lo) {
        if (lo >= 1)
          h = dp[hi] - dp[lo - 1];
        else
          h = dp[hi];
      }
      long long ans = g + h + (temp2 - temp1) * (count);
      cout << ans << " ";
    }
    cout << "\n";
  }
}
