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(){
	long long T = 1;
	cin >> T;
	while( T -- ){
		long long  a, b, q;
		cin >> a >> b >> q;
		long long mx = max(a, b), lcm = (a * b) / __gcd(a, b), _lcm = lcm;
		while( q -- ){
			long long l, r;
			cin >> l >> r; 
			l = max( mx , l);
			if( l > r ){  cout << 0 << ' '; continue; }
			long long ans = r - l + 1;
			long long l2 = (l / lcm + 1) * lcm, r2 = (r / lcm + 1) * lcm;
			long long res = (r2 - l2) / lcm ;
			if( l % lcm ) lcm = l2;
			else lcm = (l2 - _lcm), res ++;
			while( res -- ){
				if( lcm == r ){ ans --; continue; }
				 if( lcm + mx > r ){ ans -= (r - lcm) + 1; continue; }
				ans -= mx, lcm += _lcm;
			}
			cout << ans << ' ';
		}
		cout << '\n';
	}
}
