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;

int main() {
    std::ios_base::sync_with_stdio(false);
	int t;
	cin >> t;
	while(t--) {
	    long long a, b, q;
	    cin >> a >> b >> q;
	    while(q--) {
	        long long l, r;
	        cin >> l >> r;
	        long long ans = r - l + 1LL;
	        if(l <= b - 1) {
	            long long right = min(b - 1LL, r);
	            ans -= (right - l + 1LL);
	        }
	        long long lcm = (a * b) / __gcd(a, b);
	        ans -= (r / lcm);
	        ans -= ((r / lcm) - 1LL) * (b - 1LL);
	        ans -= min(b - 1LL, r - lcm * ((r / lcm) - 1LL));
	       
	        ans += ((l - 1LL) / lcm);
	        ans += (((l - 1LL) / lcm) - 1LL) * (b - 1LL);
	        ans += min(b - 1LL, l - 1LL - lcm * (((l - 1LL) / lcm) - 1LL));
	        cout << ans << " ";
	    }
	    cout << "\n";
	}
	return 0;
}
