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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modpow(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1)
      result = ((result % 1000000007) * (x % 1000000007)) % 1000000007;
    x = ((x % 1000000007) * (x % 1000000007)) % 1000000007;
    n = n / 2;
  }
  return result;
}
long long lcm(long long a, long long b) { return (a * b) / (gcd(a, b)); }
bool isprime(long long n) {
  if (n < 2) return false;
  for (long long i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return false;
  return true;
}
long long ctoi(char c) { return c - 48; }
bool visited[1000000]{false};
void add_edge(vector<long long> v[], long long x, long long y) {
  v[x].push_back(y);
  v[y].push_back(x);
}
void dfs(vector<long long> v[], long long e) {
  visited[e] = true;
  for (long long i = 0; i < v[e].size(); i++) {
    if (!visited[v[e][i]]) {
      dfs(v, v[e][i]);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    long long arr[10001]{0};
    for (long long i = 1; i <= 10000; i++) {
      if ((i % a) % b != (i % b) % a) {
        arr[i]++;
      }
    }
    for (long long i = 1; i <= 10000; i++) {
      arr[i] += arr[i - 1];
    }
    while (q--) {
      long long l, r;
      cin >> l >> r;
      long long ans = 0;
      if (l >= 10000) {
        ans = 0;
      } else {
        if (r > 10000) {
          r = 10000;
        }
        ans = arr[r] - arr[l - 1];
      }
      cout << ans << " ";
    }
    cout << "\n";
  }
  return 0;
}
