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;
typedef long long ll;
#define vi vector<int>
#define vll vector<ll>
#define pll pair<ll,ll>
#define pii pair<int,int>
#define pb  push_back
#define ppb pop_back
#define mp  make_pair
#define SIZE 201
ll m[200];
int main(){
	#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);
	ll t;
	cin>>t;
	while(t--){
		ll a,b,q;
		cin>>a>>b>>q;
		if(a>b){
			swap(a,b);
		}
		ll cnt=0;
		ll num=a*b/__gcd(a,b);
		for(ll i=0;i<a*b;++i){
			ll f=i%a;
			ll s=(i%b)%a;
			if(f!=s){
				cnt++;
			}
		}
		while(q--){
			ll l,r;
			cin>>l>>r;
			ll cnt1=(l-1)/num;
			ll cnt2=r/num;	
			ll ans=(cnt2-cnt1)*cnt;
			//cout<<ans<<",";
			for(ll i=0;i<=r%num;++i){
				ll f=i%a;
				ll s=(i%b)%a;
				if(f!=s){
					ans++;
				}
			}
			for(ll i=0;i<l%num;++i){
				ll f=i%a;
				ll s=(i%b)%a;
				if(f!=s){
					ans--;
				}
			}
			cout<<max(ans,0LL)<<" ";
		}
		cout<<endl;
	}
}