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 int gcd(long long int a, long long int b) {
  return (!b) ? a : gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int a, b, q;
long long int check(long long int x) { return ((x % a) % b) == ((x % b) % a); }
void solve() {
  cin >> a >> b >> q;
  a = min(a, b);
  b = max(a, b);
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    long long int tot = r - l + 1;
    long long int l1 = l / lcm(a, b);
    long long int r1 = r / lcm(a, b);
    long long int ans = 0;
    if (check(l)) {
      ans += (b - 1) - (l % b) + 1;
    }
    if (check(r)) {
      ans += (r % b) + 1;
    }
    ans += (r1 - l1) * b;
    cout << max(0ll, tot - ans) << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("input.txt", "r")) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int t = 1;
  int ti;
  cin >> t;
  while (t--) {
    solve();
  }
}
