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;
#define ll long long
#define ld long double
#define pb push_back
#define mpi map<ll,ll>
#define vl vector<ll>
#define mp make_pair
#define pl pair<ll,ll>
#define forn(i,n) for(ll i=0;i<n;i++)
#define ff first
#define ss second
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define inf       1000000000000
#define MOD       1000000007


void solve()
{
    ll a,b,q;
    cin>>a>>b>>q;
    ll g=__gcd(a,b);
    ll l=(a*b)/g;
    forn(i,q)
    {
    	ll x,y;
    	cin>>x>>y;
        ll ans=0;
        ll t1=x/l,t2=y/l;
        if(y<max(a,b))
        {
        	cout<<0<<" ";
        	continue;
        }
        ans=max(l*t1-x,(ll)0)+max(y-(l*t2+max(a,b))+1,(ll)0);
        ans+=max(min(l*t2-1,y)-max(l*t1+max(a,b),x)+1,(ll) 0)-max(a,b)*(max(t2-t1-1,(ll)0));
        cout<<ans<<" ";
    }
}

int main()
{
	fast
	//pre();
	ll t=1;
	cin>>t;
	while(t--)
	{
		solve();
		cout<<endl;
	}
}   