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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
long long mod = 1e9 + 7;
const int MAXN = 10005;
const long double INF = 1e18 + 9;
int s = 1024;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a * b / gcd(a, b); }
int32_t main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int a, b, q;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    int c = lcm(a, b);
    while (q--) {
      long long l, r;
      cin >> l >> r;
      int l1 = l % c;
      l += c - l1;
      int r1 = r % c;
      r -= r1;
      long long res = 0;
      if (l <= r) {
        long long k = r / c - l / c;
        res += k * (c - b);
        res += c - max(b, l1);
        res += max(0, r1 - b + 1);
      } else {
        l1 = max(l1, b);
        if (l1 <= r1) res += r1 - l1 + 1;
      }
      cout << res << ' ';
    }
    cout << endl;
  }
  return 0;
}
