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;
struct modll {
  long long num;
  operator long long() { return num; }
};
modll operator*(modll x, modll y) { return {(x.num * y.num) % 1000000007}; }
modll operator+(modll x, modll y) {
  return {(x.num + y.num + 1000000007) % 1000000007};
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int tt;
  cin >> tt;
  while (tt--) {
    int a, b, q;
    cin >> a >> b >> q;
    int prod = (a * b);
    long long l, r, l_dup;
    while (q--) {
      cin >> l >> r;
      l_dup = l;
      long long mini = min(a, b), maxi = max(a, b);
      long long prod = a * b;
      prod /= gcd(a, b);
      if (maxi == prod) {
        cout << 0 << ' ';
        continue;
      }
      long long ans = 0;
      while (l % prod <= maxi && l <= r) {
        ans++;
        l++;
        if (l % prod == maxi) {
          l++;
        }
      }
      long long low = (l - 1) / prod, high = r / prod;
      long long num = high - low;
      ans += num * maxi;
      if (r - prod * high + 1 < maxi) ans -= (maxi - (r - prod * high + 1));
      cout << (r - l_dup + 1) - ans << ' ';
    }
    cout << '\n';
  }
  return 0;
}
