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;
typedef long long ll;
#define f(i,a,b) for(ll i=(a);i<(b);i++)
#define frr(i,a,b) for(ll i=(a-1);i>=(b);i--)
#define vi vector<int>
#define de deque<int>
#define del deque<ll>
#define vl vector<ll>
#define pb push_back
#define ppb pop_back()
#define pf push_front
#define ppf pop_front()
#define mpp map<ll,ll>
#define rev(a) reverse(a.begin(),a.end())
#define srt(a) sort(a.begin(),a.end())
#define ln(a) a.length()
#define sz(a) a.size()
#define endl "\n"
int M=1000000007;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);

   int t;cin>>t;
   while(t--)
   {
       int a,b,q;
       cin>>a>>b>>q;if(a>b)swap(a,b);
       int div=(a*b)/__gcd(a,b);
       while(q--)
       {
           ll l,r;cin>>l>>r;
           if(l==r || a==b)cout<<0<<" ";
           else{ll k=r/div;
           ll dif1=0;
           if(r>=(div*(k))+(b-1)){dif1+=k*(b-1);}
           else{dif1+=(k-1)*(b-1)+(r-(k*div));}
           ll k1=l/div;
           ll dif2=0;
           if(l>=(div*(k1))+(b-1)){dif2+=k1*(b-1);}
           else{dif2+=(k1-1)*(b-1)+(l-(k1*div));}
           //cout<<dif1<<" "<<dif2<<endl;
           cout<<(r-l)-(dif1-dif2)<<" ";}

       }
       cout<<endl;
   }

}

