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 gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(int a, int b) { return (a * b) / gcd(a, b); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long high = max(a, b);
    long long low = min(a, b);
    long long p = lcm(a, b);
    while (q--) {
      long long l, r;
      cin >> l >> r;
      if (high % low == 0) {
        cout << 0 << " ";
        continue;
      }
      if (r < b) {
        cout << 0 << " ";
        continue;
      }
      l = max(b, l);
      long long x = l;
      while (x < r) {
        if (x % p == 0) break;
        x++;
      }
      long long y = r;
      while (y > l) {
        if (y % p == 0) break;
        y--;
      }
      long long ans = 0;
      if (x == y) {
        if ((x + high) < r)
          ans += high;
        else
          ans += (r - x + 1);
      } else if (x < y) {
        long long g = x / p;
        long long h = y / p;
        long long count = h - g;
        ans += (count * high);
        if ((y + high) < r)
          ans += high;
        else
          ans += (r - y + 1);
      }
      cout << r - l + 1 - ans << " ";
    }
    cout << endl;
  }
  return 0;
}
