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 main(){
	int t;
	cin>>t;
	while(t--){
		int a,b,q;
		cin>>a>>b>>q;
		int res[a*b]{};
		int count=0;
		for(int i=0;i<a*b;i++){
			if((i%a)%b==(i%b)%a){
				res[i]=0;
			}else{
				res[i]=1;
				count++;
			}
		}
		for(int i=0;i<q;i++){
			long long int l,r;
			cin>>l>>r;
			long long int ans=0;
			while(l%(a*b)!=0 && l<r){
				ans+=res[l%(a*b)];
				l++;
			}
			while(r%(a*b)!=0 && r>l){
				ans+=res[r%(a*b)];
				r--;
			}
			ans+=((r-l+1)/(a*b))*count;
			cout<<ans<<endl;
		}
	}
}