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 cnt[201];
long long upb(long long x, long long y) {
  long long res = (x / y) * y;
  return res == x ? res : res + y;
}
long long lwb(long long x, long long y) {
  long long res = (x / y) * y;
  return res == (x - y + 1) ? x + 1 : res;
}
void run() {
  int a, b, q;
  cin >> a >> b >> q;
  int c = a * b;
  int sum = 0;
  memset(cnt, 0, b * sizeof(int));
  for (int i = 0; i < c; ++i) cnt[i / a] += (i % a % b != i % b % a);
  for (int i = 0; i < b; ++i) sum += cnt[i];
  for (int z = 0; z < q; ++z) {
    long long l, r;
    long long ans = 0;
    cin >> l >> r;
    long long k = upb(l, c), p = lwb(r, c);
    if (k > r) {
      long long g = upb(l, a);
      if (g > r) {
        for (long long i = l; i <= r; ++i) ans += (i % a % b != i % b % a);
      } else {
        for (long long i = l; i < g; ++i) ans += (i % a % b != i % b % a);
        for (; g + a - 1 <= r; g += a) ans += cnt[(g / a) % b];
        for (; g <= r; ++g) ans += (g % a % b != g % b % a);
      }
    } else {
      long long g = upb(l, a);
      for (long long i = l; i < g; ++i) ans += (i % a % b != i % b % a);
      for (; g + a <= k; g += a) ans += cnt[(g / a) % b];
      ans += ((p - k) / c) * sum;
      for (; p + a - 1 <= r; p += a) ans += cnt[(p / a) % b];
      for (; p <= r; ++p) ans += (p % a % b != p % b % a);
    }
    cout << ans << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) run();
  return 0;
}
