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<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<math.h>
#include<map>
typedef long long int ll;
using namespace std;
#define maxn 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
const int mm=1e6;
ll d[mm];
int main()
{
	ll n,m,i,j,k,kk,a,b,t,c,nn,mm,s,ss,nnn,mmm;
	cin>>t;
	while(t--)
	{
		cin>>a>>b>>c;
		if(a>b)
		swap(a,b);
		for(i=0;i<c;i++)
		{
		   	k=a*b/__gcd(a,b);
//		    cout<<k<<endl;
		    s=0; ss=0;
			cin>>n>>m;
			n--;
			nn=n/k;
			s=nn*(k-b);
			s+=max(n%k-b,(ll)0);
			mm=m/k;
			ss=mm*(k-b);
			ss+=max(m%k-b,(ll)0);
//			cout<<ss<<' '<<s<<endl;
            if(m>=b&&n<b)
            ss++;
			cout<<ss-s<<endl;
		 } 
	}

}