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;
int gcd(int a, int b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int main() {
  int i, t, q, j, a, b, c, lll;
  long long int l, r, la, rb, dl, dr, d, s;
  long long int sum, ans;
  cin >> t;
  while (t--) {
    lll = 0;
    scanf("%d%d%d", &a, &b, &q);
    c = a * b / gcd(a, b);
    if (b % a == 0 || a % b == 0) lll = 1;
    for (i = 1; i <= q; i++) {
      cin >> l >> r;
      if (lll == 1)
        cout << '0' << ' ';
      else {
        sum = 0;
        la = l;
        la /= c;
        la *= c;
        rb = r;
        rb /= c;
        rb *= c;
        dl = l - la;
        dr = r - rb;
        if (la == rb) {
          if (dl < b) {
            if (dr >= b)
              sum += b - dl;
            else {
              sum += r;
              sum -= l;
              sum++;
            }
          }
          s = r;
          s -= l;
          s++;
          s -= sum;
          cout << s << ' ';
        } else {
          if (dl < b) sum += b - dl;
          if (dr < b) {
            sum += dr + 1;
            d = rb;
            d -= c;
          } else
            d = rb;
          d -= la;
          d /= c;
          ans = d;
          ans *= b;
          sum += ans;
          ans = r;
          ans -= l;
          ans++;
          ans -= sum;
          cout << ans << ' ';
        }
      }
    }
    cout << endl;
  }
  return 0;
}
