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
ll count(ll r,ll b,ll lcm){
   ll div=r/lcm;
   ll prod=div*lcm;
   ll res=r;
   if(div!=0){
      res-=min(b,r-div+1);
      res-=(div-1)*b+min(b-1,r);}
      
      return res;
   
   }
int main(){
   int t;cin>>t;
   while(t--){
      ll a,b,q;cin>>a>>b>>q;
      a=min(a,b);
      b=max(a,b);
      ll lcm=a*b/__gcd(a,b);
      while(q--){
         ll l,r,ans=0;
         cin>>l>>r;
         if ((lcm == b) || (r < b) ) {
                cout << 0 << " ";
                continue;
            }
            else{
         ans+=count(r,b,lcm)-count(l-1,b,lcm);
         cout<<ans;
      }
      }
      cout<<endl;
   }
   }
