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 mp make_pair
#define ff first
#define ss second
#define pb push_back
#define w(x) scanf("%d",&x);while(x--)
#define rep(i,a,b) for(i=a;i<=b;i++)
#define ios ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pi 2*acos(0.0)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef vector<int> vi;

long long int mod=1e9+9;
const int N=100000;
long long l,r, t,i,j,cs=0,n,a,b;
//string s;
long long lcm(ll a,ll b)
{
    ll g=__gcd(a,b);
    if(b>a) return (b/g) * a;
    return (a/g)*b;
}
int main()
{
    //ios
    w(t){
        cin>>a>>b>>n;
        ll m=max(a,b);
        //ll s=min(a,b);
        ll lc=lcm(a,b);
        while(n--){
            cin>>l>>r;
            if(m> l && m> r){
                cout<<0<<" ";
                continue;
            }
            if(l<m) l=m;
            long long ans=r-l+1;

            ll f=(l/lc);
            ll sec=r/lc;
            ans-=(sec-f)*m;
         //   if(sec==f) ans-=m;
            //cout<<ans<<" "<<0;
//            if( f*lc<l && f*lc+m>l){
//                ans+=(l-f*lc);
//            //     cout<<ans<<" "<<1;
//
//            }
            if(sec!=0 && sec*lc+m>=r){
                ans+=sec*lc+m-r;
             //    cout<<ans<<" "<<2;
             //if(sec*lc+m==r) ans--;
            }

            cout<<ans<<" ";

        }
        cout<<endl;
    }
}


