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::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t, a, b, q, i;
  long long int tmp, j;
  cin >> t;
  while (t--) {
    cin >> a >> b >> q;
    vector<vector<long long int>> v(q, vector<long long int>(2, 0));
    vector<long long int> ans(q, 0);
    for (i = 0; i < q; i++) {
      cin >> v[i][0] >> v[i][1];
    }
    for (i = 0; i < q; i++) {
      for (j = i + 1; j < q; j++) {
        if (v[j][0] < v[i][0] || (v[i][0] == v[j][0] && v[j][1] < v[i][1])) {
          tmp = v[i][0];
          v[i][0] = v[j][0];
          v[j][0] = tmp;
          tmp = v[i][1];
          v[i][1] = v[j][1];
          v[j][1] = tmp;
        }
      }
    }
    for (j = v[0][0]; j <= v[0][1]; j++) {
      if ((j % a) % b != (j % b) % a) {
        ans[0]++;
      }
    }
    for (i = 1; i < q; i++) {
      if (v[i][0] == v[i - 1][0]) {
        ans[i] = ans[i - 1];
        if (v[i][1] != v[i - 1][1]) {
          for (j = v[i - 1][1] + 1; j <= v[i][1]; j++) {
            if ((j % a) % b != (j % b) % a) {
              ans[i]++;
            }
          }
        }
      } else {
        if (v[i][0] <= v[i - 1][1] && v[i - 1][1] <= v[i][1]) {
          ans[i] = ans[i - 1];
          for (j = v[i - 1][0]; j < v[i][0]; j++) {
            if ((j % a) % b != (j % b) % a) {
              ans[i]--;
            }
          }
          for (j = v[i - 1][1] + 1; j <= v[i][1]; j++) {
            if ((j % a) % b != (j % b) % a) {
              ans[i]++;
            }
          }
        } else {
          for (j = v[i][0]; j <= v[i][1]; j++) {
            if ((j % a) % b != (j % b) % a) {
              ans[i]++;
            }
          }
        }
      }
    }
    for (i = 0; i < q; i++) {
      cout << ans[i] << " ";
    }
    cout << '\n';
  }
  return 0;
}
