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;
constexpr int inf = 0x3f3f3f3f;
constexpr long long INF = 9123456789123456789;
constexpr long long MOD = 1000000007LL;
unsigned int gcd(unsigned int a, unsigned int b) {
  while (b != 0) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
unsigned int lcm(unsigned int a, unsigned int b) { return a * b / gcd(a, b); }
unsigned int a, b, q;
void sol() {
  cin >> a >> b >> q;
  if (a > b) swap(a, b);
  unsigned int period = lcm(a, b);
  vector<int> mark(period);
  unsigned int cnt(0);
  for (unsigned int i = 0; i < period; i++) {
    if (i < b)
      mark[i] = cnt;
    else
      mark[i] = (++cnt);
  }
  for (unsigned int i = 0; i < q; i++) {
    long long int l, r;
    cin >> l >> r;
    l--;
    long long int rn = (r / period) * cnt + mark[r % period];
    long long int ln = (l / period) * cnt + mark[l % period];
    cout << (rn - ln) << ' ';
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(12);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) sol();
  return 0;
}
