Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long int nxtl() {
  long long int x;
  cin >> x;
  return x;
}
void SieveOfEratosthenes(int n, vector<int> &primes) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int power(long long int x, long long int y, long long int mod) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}
long long int cal(vector<long long int> &prefix, long long int a,
                  long long int b, long long int x);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = nxt();
  while (t--) {
    long long int a = nxtl(), b = nxt(), q = nxtl();
    vector<int> v(a * b + 1, 0);
    for (int i = 1; i <= a * b; i++) {
      if ((i % a) % b != (i % b) % a) v[i] = 1;
    }
    vector<long long int> prefix(a * b + 1);
    prefix[0] = 0;
    for (int i = 1; i <= a * b; i++) prefix[i] = prefix[i - 1] + v[i];
    while (q--) {
      long long int test = 0;
      long long int l = nxtl(), r = nxtl();
      test = cal(prefix, a, b, r) - cal(prefix, a, b, l - 1);
      cout << test << " ";
    }
    cout << endl;
  }
  return 0;
}
long long int cal(vector<long long int> &prefix, long long int a,
                  long long int b, long long int x) {
  long long int ret = 0;
  ret = prefix[a * b] * (x / (a * b));
  long long int rem = x % (a * b);
  ret += prefix[rem];
  return ret;
}