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;
#define forr(i,a,n)	for(long long int i=a; i<n; i++)
#define loop(i,a,n)	for(long long int i=a; i>=n; i--)

long long int lcm(long long int a,long long int b)	{long long int x=((1ll*a*b)/__gcd(a,b)); return x;}
long long int check(long long int l, long long int r, long long int x,long long int a,long long int p)
{
	long long int ans=(r-p*x)/(x*1ll)+1ll;
	return ans;
}
int main() 
{
	ios_base :: sync_with_stdio(false),cin.tie(NULL),cout.tie(0);

	int m; cin>>m; while(m--)
	{
		long long int a,b,c,ct=0;
		cin>>a>>b>>c;
		if(a>b)
			swap(a,b);
		long long int l,r,x=lcm(a,b),p,ans=0;
		//cout<<x<<'\n';
		forr(i,0,c)
		{
			cin>>l>>r;
			ans=r-l+1;
			p=ceil(double(l)/x);
			//cout<<ans<<'\n';
			if(l<b)	ans-=b-l;
			//cout<<ans<<'\n';
			if(x>r)
			{
				cout<<max(0ll,ans)<<' ';
				continue;
			}
			long long int last=check(l,r,x,a,p);
			//cout<<ans<<' '<<last<<' '<<p*x<<' '<<last*x<<' '<<p<<'\n';
			long long int back=(p-1)*x; 
			back+=b-1;
			//cout<<back<<'\n';
			if(back>=l)
				ans-=abs(back-l+1);
			//cout<<ans<<'\n';
			if(p*x<=r)	ans-=last;
			//cout<<ans<<'\n';
			long long int la=(last+p-1)*x;
			//cout<<la<<'\n';
			if(la+b-1<=r)
				ans-=last*(b-1);
			else
			{
				ans-=(last-1)*(b-1);
				if(la<r)
				ans-=abs(la-r+1);
			}
			cout<<max(0ll,ans)<<' ';
		}
		cout<<'\n';
	}
	
	return 0;
}