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) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  long long a, b, q;
  long long l, r;
  int t;
  long long ans;
  scanf("%d", &t);
  while (t--) {
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    long long lcm = b / gcd(a, b) * a, tmp, num;
    while (q--) {
      scanf("%I64d%I64d", &l, &r);
      if (a == b || b % a == 0)
        printf("0\n");
      else {
        ans = (r - l + 1) / lcm * (lcm - b);
        num = (r - l + 1) % lcm;
        if (num > 0) {
          tmp = l % lcm;
          if (tmp + num - 1 >= b)
            ans += min(lcm - 1, tmp + num - 1) - max(b, tmp) + 1;
          if (tmp + num - 1 >= lcm) {
            num -= lcm - tmp;
            tmp = 0;
            ans += max(0LL, num - b);
          }
        }
        printf("%I64d%c", ans, q ? ' ' : '\n');
      }
    }
  }
  return 0;
}
