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;
const double pi = acos(-1);
const long long mod = 1e9 + 7;
long long s[1000];
long long gcd(long long x, long long y) {
  if (x % y == 0)
    return y;
  else {
    return gcd(y, x % y);
  }
};
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(s, 0, sizeof(s));
    long long a, b, q, l, r;
    cin >> a >> b >> q;
    long long gg = gcd(a, b);
    long long ss = b / gg * a;
    for (int i = 0; i < q; i++) {
      cin >> l >> r;
      if (a == b) {
        s[i] = 0;
        continue;
      }
      if (r < b)
        s[i] = 0;
      else {
        long long sum = r - l + 1;
        sum -= ((r / ss) * b);
        if (r / ss * ss + b >= r) {
          sum += (r / ss * ss + b - r - 1);
        }
        sum += ((l / ss) * b);
        if ((l) / ss * ss + b >= l) {
          sum -= (l / ss * ss + b - l);
        }
        s[i] = sum;
      }
    }
    for (int i = 0; i < q; i++) {
      if (i == 0)
        cout << s[i];
      else
        cout << " " << s[i];
    }
    cout << endl;
  }
  return 0;
}
