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 int long long
#define rep(i,x,y) for(int i=x;i<y;i++)
#define repr(i,x,y) for(int i=x;i>=y;i--) 
#define mod 1000000007
using namespace std;

void Onigiri()
{
  int a,b,q;
  cin>>a>>b>>q;
  if(a>b)swap(a,b);
  int g=__gcd(a,b);
  int lcm=a*b/g;
  while(q--)
  {
    int l,r;
    cin>>l>>r;
    int a1=0,a2=0;
    l--;
    int lo=0,hi=ceil(1.00*r/lcm);
    while(lo<=hi)
    {
      int mid=(lo+hi)/2;
      if(mid*lcm<=r)
      {
        lo=mid+1;
        if(lcm*(mid+1)>=r)
        {
          a1=mid*b+min(r-mid*lcm+1,b);
          break;
        }
      }
      else hi=mid-1;
    }
   lo=0,hi=ceil(1.00*l/lcm);
   while(lo<=hi)
    {
      int mid=(lo+hi)/2;
      if(mid*lcm<=l)
      {
        lo=mid+1;
        if(lcm*(mid+1)>=l)
        {
          a2=mid*b+min(l-mid*lcm+1,b);
          break;
        }
      }
      else hi=mid-1;
    }
    cout<<r-l-a1+a2<<" ";
  }

}
signed main()
{
   ios_base::sync_with_stdio(false);cin.tie(NULL);
   #ifdef Zoro
   freopen("in", "r", stdin);freopen("out", "w", stdout);
   #endif  
 
   int t=1; 
   cin>>t;
 
   while(t--)
   {Onigiri();cout<<"\n";}

   cerr<<"\n"<<(float)clock()/CLOCKS_PER_SEC*1000<<" ms"<<endl;
   return 0;
}