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>
const long long INF = LLONG_MAX;
const long long MOD = 1e9 + 7;
using namespace std;
void fastio() {}
long long gcd(long long r, long long b) {
  long long x = 1;
  while (b != 0) {
    x = b;
    b = r % b;
    r = x;
  }
  return x;
}
void solve() {
  long long a, b, q, z = 0;
  cin >> a >> b >> q;
  long long c = gcd(a, b);
  c = (a * b) / c;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long d = (r / c) - (l / c);
    long long ans = r - l + 1;
    if (l >= c) {
      for (long long i = (l / c) * c; i < (l / c) * c + max(a, b); i++) {
        if (l <= i) ans--;
      }
      for (long long i = (r / c) * c; i < (r / c) * c + max(a, b); i++) {
        if (r < i) ans++;
      }
    }
    ans = ans - d * max(a, b);
    if (b > l) {
      ans -= b - l;
    }
    cout << max(ans, z) << ' ';
  }
  cout << endl;
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
