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>
#define ll          long long
#define pb          push_back
#define	endl		'\n'
#define pii         pair<ll int,ll int>
#define vi          vector<ll int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll int)x.size()
#define hell        1000000000000000000
#define rep(i,a,b)	for(ll int i=a;i<b;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
// #define mp          make_pair
#define MAX 200005 
#define CHAR_SIZE 26
using namespace std;

#define N  1500001



void solve(ll cno)
{
	ll n,i,j,k,l,r,m,q,h;
	ll a,b;
	cin>>a>>b>>q;

	ll lcm=(a*b)/(__gcd(a,b));
		// cout<<lcm<<endl;
	ll arr[lcm+1];
	ll cnt=0;
	arr[0]=0;
	
	for(i=1;i<=lcm;i++)
	{
		if(((i%a)%b)==((i%b)%a))
			arr[i]=arr[i-1];
		else
			{arr[i]=arr[i-1]+1;
				cnt++;}
	}
	// cout<<cnt<<endl;
	while(q--)
	{

		cin>>l>>r;
		if(a==b)
		{
			cout<<0<<" ";
		}
		l=l-1;
		ll d=l/lcm;
		ll left=l%lcm;
		ll ans1=d*arr[lcm]+arr[left];
		// cout<<ans1<<" ";
		ll d2=r/lcm;
		ll left2=r%lcm;
		ll ans2=d2*arr[lcm]+arr[left2];

		cout<<ans2-ans1<<" ";
	}

cout<<endl;


}
// cout<<mn<<endl;
/*

*/


int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int TESTS=1;
	cin>>TESTS;
	ll i=1;
	while(TESTS--)
	{
		solve(i);
		i++;
	}
	return 0;
} 