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;
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
void out(vector<long long> arr) {
  for (auto i : arr) {
    cout << i << ' ';
  }
  cout << "\n";
}
void out(vector<pair<long long, long long>> par) {
  for (auto i : par) {
    cout << i.first << ' ';
  }
  cout << endl;
  for (auto i : par) {
    cout << i.second << ' ';
  }
  cout << endl;
}
void out(set<long long> arr) {
  for (auto i : arr) {
    cout << i << ' ';
  }
  cout << endl;
}
void inp(vector<long long> &arr) {
  for (long long i = 0; i < arr.size(); i++) {
    cin >> arr[i];
  }
}
void inp(long long &n) { cin >> n; }
void inp(long long &a, long long &b, long long &c) { cin >> a >> b >> c; }
void inp(long long &a, long long &b) { cin >> a >> b; }
void inp(long long &a, long long &b, long long &c, long long &d) {
  cin >> a >> b >> c >> d;
}
long long MOD = 1e9 + 7;
const long long N = 2 * 100000 + 10;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    long long gg = gcd(a, b);
    gg = (a * b) / gg;
    for (long long i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      long long v1 = l / gg;
      long long v2 = r / gg;
      long long ans = (v2 - v1 + 1) * b;
      v1 = l - v1 * gg;
      v1 = min(b, v1);
      v2 = v2 * gg + b - 1 - (r);
      v2 = max(0, v2);
      ans = ans - v1 - v2;
      cout << (r - l + 1) - ans << "\n";
    }
  }
}