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 ll long long int
#define pb push_back
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mp make_pair 
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define pre(i , j , n) for(ll i = j ; i >= n ; i--) 
#define vi vector<int>
#define vll vector<ll> 
#define inf 9223372036854775806
#define mod 1000000007
#define brl cout<<"\n"
using namespace std;
void fast()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
void inout()
{
  #ifndef ONLINE_JUDGE  
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  #endif
}
int main() 
{
  fast();
  inout();
  //code goes here
  int t;
  cin>>t;
  while(t--)
  {
    int a,b,q;
    cin>>a>>b>>q;
    if(a>b)swap(a,b);
    int lcm=(a*b)/(__gcd(a,b));
    while(q--)
    {
      ll l,r;
      cin>>l>>r;
      if(b%a==0){
        cout<<0<<" ";
        continue;
      }
      if(r<=b){
        cout<<0<<" ";
        continue;
      }
      if(r<=lcm){
        ll ans=r-max(l,(ll)b)+1;
        if(r==lcm)ans--;
        cout<<ans<<" ";
        continue;
      }
     ll lower=ceil((double)l/lcm);
     ll upper=r/lcm;
     ll ans=(max((ll)0,upper-lower))*b;
     upper=upper*lcm;
     ans+=min(r+1,upper+b)-upper;
     cout<<r-l+1-ans<<" ";
   }
    brl;
  }
}