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
#define mkp make_pair
#define pb push_back
#define ld long double

int main() {
    ll t;
    cin>>t;
    while(t--){
       ll a, b, q;
       cin>>a>>b>>q;
       ll gcd = __gcd(a, b);
       ll lcm = (a*b)/gcd;
       while(q--){
           ll l, r;
           cin>>l>>r;
           if((l < a && r < a)||  (r < b)){
               cout<<0<<" ";
           } else{
               /*ll c=0;
               for(ll i = l;i<=r;i++){
                   if((i%a)%b == (i%b)%a){
                       //cout<<i<<endl;
                       c++;
                   }
               }*/
               //cout<<"ans1="<<((r-l+1)-c)<<endl;
               ll mn = l/lcm;
               ll mx = r/lcm;
               ll ans = (r-max(max(a,b),l)+1)-((mx-mn)*max(a,b));
               cout<<ans<<" ";               
           }
       }
        cout<<endl;
    }
}
