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(long long a, long long b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    vector<long long> p;
    long long a, b, q;
    cin >> a >> b >> q;
    long long l, r;
    if (a == b) {
      while (q--) {
        cin >> l >> r;
        p.push_back(0);
      }
    } else {
      long long u = a * b / gcd(a, b);
      long long v;
      if (a > b)
        v = a;
      else
        v = b;
      if (v % (a + b - v) == 0) v++;
      v--;
      while (q--) {
        cin >> l >> r;
        l--;
        if (l < v) l = v;
        if (r <= v) {
          p.push_back(0);
          continue;
        }
        long long x = r - l;
        x -= (r / u - l / u) * (v + 1);
        long long c = r - r / u * u;
        if (c < v && r >= u) x += v - c;
        p.push_back(x);
      }
    }
    for (long long i = 0; i < p.size(); i++) {
      if (i) cout << " ";
      cout << p[i];
    }
    cout << endl;
  }
  return 0;
}
