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;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long int INFLL = 0x3f3f3f3f3f3f3f3fLL;
const long double PI = 3.141592653589793;
const int MAX = 4e4 + 5;
int v[MAX], n;
void build(int a, int b) {
  n = a * b;
  v[0] = 0;
  for (int i = 1; i <= n; ++i) {
    v[i] = v[i - 1] + (((i % a) % b) != ((i % b) % a));
  }
}
long long int query(long long int p) {
  long long int aux = p / n;
  int rem = p % n;
  return v[n] * 1LL * aux + v[rem];
}
long long int query(long long int l, long long int r) {
  return query(r) - query(l - 1);
}
void solve() {
  int a, b, q;
  cin >> a >> b >> q;
  build(a, b);
  for (int i = 0; i < q; ++i) {
    long long int l, r;
    cin >> l >> r;
    cout << query(l, r) << " \n"[i == q - 1];
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
