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 lcm(int a, int b)  
 {  
    return (a*b)/__gcd(a, b);  
 }

int main()
{
	long long int ans, a, b, t, q, l, r, i, j, k, c, d, p;
	cin>>t;
	while(t--) {
		cin>>a>>b>>q;
		k=lcm(a, b);
		while(q--) {
			cin>>l>>r; c=0;
			i=max(l, max(a, b));
			if(i<=r) {
				for(j=i; j<=r; j++) {
					if(j%k==0) { c++; break; }
				}
				if(c) {
					d=r-j;
					c+=d/k; c--; c*=max(a, b);
					p=d%k; c+=min(p, max(a, b));
				}
				ans=(r-i+1)-c;
			}
			else ans=0;
			cout<<ans<<" ";
		}
		cout<<endl;
	}
}