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 int long long
#define mod 1000000007
#define pf push_forward
#define pb push_back
#define mp make_pair
#define f(i,in,n) for(i=in; i<n; ++i)
#define scant int t; cin>>t;while(t--)
using namespace std;
main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    scant{
    	int a,b,q;
    	cin>>a>>b>>q;
    	int mul=a*b;
    	while(q--){
    		int l,r,i;
    		cin>>l>>r;
    		if(a==b){
    			cout<<0<<" ";
    			continue;
    		}
    		l=max(l,b);
    		if(l>r){
    			cout<<0<<" ";
    			continue;
    		}
    		int ans = r-l+1;
    		int count=1;
    		while(mul<=r){
    			mul = count*mul;
    			if(mul<=r&&mul>=l){
    				if(mul+9<=r){
    					ans-=10;
    				}
    				else{
    					ans-=(r-mul+1);
    				}
    			}
    			count++;
    		}
    		cout<<ans<<" ";
    	}
    	cout<<endl;
    }
}