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 N = (int)2e5 + 123;
const int K = 500;
const int maxn = (int)2e9;
const int mod = 100000007;
const long long inf = 1ll * 1e18 + 123;
int a, b, qq;
int dp[40022];
long long solve2(long long p) {
  long long t = p / (a * b);
  int k = p % (a * b);
  return 1ll * dp[a * b] * t + dp[k];
}
void solve() {
  cin >> a >> b >> qq;
  for (int i = 1; i <= a * b; ++i) {
    dp[i] = dp[i - 1];
    if ((i % a) % b != (i % b) % a) {
      dp[i]++;
    }
  }
  while (qq--) {
    long long l, r;
    cin >> l >> r;
    cout << solve2(r) - solve2(l - 1) << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int q;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    solve();
  }
  return 0;
}
