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>
	//#include <boost/math/common_factor.hpp>
    #define ll          long long int
    #define ld           long double
    #define ull         unsigned long long int
    #define fast() ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
    #define mem(a,val)  memset(a,val, sizeof(a) );
    #define max(a, b) (a < b ? b : a)
    #define min(a, b) (a > b ? b : a)
	#define ln '\n'
    const ll mod=1e12;
    const unsigned int N = 1e6+1;
    ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    using namespace std;
    int main(){
        fast();
        int t;
        cin>>t;
        
        while(t--){
        	int a,b,q;
        	cin>>a>>b>>q;
        	while(q--){
        		ll l,r;
        		cin>>l>>r;
        		ll ans;
        		ll mx=max(a,b);
        		if(a<b)swap(a,b);
        		if(r<mx)cout<<0<<" ";
        		else if(a%b==0)cout<<0<<" ";
        		else{
        			ll k=(a*b)/__gcd(a,b);
        			if(l<mx){
        				ans=r-mx+1;

        			}
        			else ans=r-l+1;
        			//ll k=boost::math::lcm(a,b);
        			ll c;
        			//cout<<ln<<ans<<ln;
        			if(l<=k){
        				c=r/k;
        				if(l==r && r%k==0)ans=0;
        				else{
        					if(c*k+mx<=r)ans=ans-(c*mx);
        					else ans=ans-((c-1)*mx)-(r-c*k+1);
        				}
        			}
        			else{
        				c=r/k-l/k;
        				if(l==r && r%k==0)ans=0;
        				else{
        					if((c+l/k)*k+mx<=r)ans=ans-(c*mx);
        					else ans=ans-(((c-1)*mx)+(r-(c+l/k)*k+1));
        				}
        				//cout<<ln<<(c-1)*mx<<" "<<r-(c+1)*k+1<<ln;
        				//cout<<ln<<(c+1)*k<<ln;

        			}
        			
        			cout<<ans<<" ";//<<k<<" "<<c<<" ";
        	}
        }
        	cout<<ln;
        }
        return 0;
    }
