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
 
int main()
{
    //freopen("in.txt","r",stdin);
    ll t;
    cin>>t;
    while(t--)
    {
        ll a,b,q,l,r;
        cin>>a>>b>>q;
        if(a>b)swap(a,b);
        ll x=__gcd(a,b);
        x=a*b/x;
        for(ll i=0;i<q;i++)
        {
            cin>>l>>r;
            if(a==b || r<b){cout<<0<<" ";continue;}
            if(a==1)
            {
                ll m=(r/b)-(l/b);
                if(l%b==0)m++;
                cout<<m<<" ";continue;
            }
            if(l<b)l=b;
            if(x>r){cout<<r-l+1<<" ";continue;}
            ll y=ceil((double)l/x),z=floor((double)r/x);
            ll v=z-y+1;
            if((x*z)+b>r){cout<<(r-l+1)-(v*(b))+((x*z)+b-r-1)<<" ";continue;}
            cout<<(r-l+1)-(v*(b))<<" ";
        }
        cout<<endl;
    }
    return 0;
}