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 
/*
Suraj Goel
MNNIT
*/
#include<bits/stdc++.h>
#define linf 1000000000000000000
using namespace std;
typedef long long int ll;
ll modpower(ll a,ll b,ll c){
	ll res=1;
	while(b)
	{
		if(b&1LL)
			res=(res*a)%c;
		a=(a*a)%c;
		b>>=1;
	}
	return res;

}	


int main() {
    int t;
    cin >> t;
    while(t--)
    {
        ll myans=0,a,b,q,lcm;
        
 
        cin >> a >> b >> q;
        ll hcf=__gcd(a,b);
        lcm=(a/hcf)*b;
        
        while(q--)
        {
        	myans=0;
        	ll l,r,p,q,x,y,z,ank;
        	cin >> l >> r;
        	if(r<max(a,b))
        	{
        		cout << "0 ";
        		continue;
			}
        	if(l%lcm==0)
        	{
        		y=(r/lcm)*lcm;
				ank=l;
				myans+=(0ll,((y-ank)/lcm)*max(a,b));
        		if(r%lcm!=0)
				{
					
					myans+=min(max(a,b),r-y+1);
 
				}
				else
				{
				
					myans++;
				}	
			}
			else
			{
				x=(l/lcm)*lcm;
				x+=max(a,b);
				if(x>r)
				{
					x=r;
					myans+=max(0ll,x-l+1);
					cout << myans << " ";
					continue;
 
				}
				myans+=max(0ll,x-l);
 
				y=(r/lcm)*lcm;
					ank=lcm*((l/lcm)+1);
					myans+=(0ll,((y-ank)/lcm)*max(a,b));
 
				if(r%lcm!=0)
				{
					
					myans+=min(max(a,b),r-y+1);
 
				}
				else
				{
				
					myans++;
				}	
				
 
			}
			cout << max(0ll,(r-l+1-myans)) << " ";
		}
		cout << "\n";
    }
	return 0;
}