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;
typedef long double ld;
template <typename X> ostream& operator << (ostream& x,const vector<X>& v){for(ll i=0;i<v.size();++i) x<<v[i]<<" ";return x;}
template <typename X> ostream& operator << (ostream& x,const set<X>& v){for(auto it:v) x<<it<<" ";return x;}
template <typename X, typename Y> ostream& operator << (ostream& x, const pair<X,Y>& v) {x<<v.ff<<" "<<v.ss;return x;}
template <typename T, typename S> ostream& operator << (ostream& os, const map<T, S>& v) { for (auto it : v) os << it.first << "=>" << it.second << endl; return os; }
struct pair_hash {inline std::size_t operator()(const std::pair<ll, ll> & v) const {return v.first*31+v.second;}};
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC targett("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define speed ios_base::sync_with_stdio(false);cin.tie(NULL)
#define pb push_back
#define fr first
#define sc second
#define prnt(m) for(auto it = m.begin() ; it != m.end() ; ++it){cout<<*it<<endl;}
#define tra(it,m) for(auto it=m.begin();it!=m.end();++it)
#define DBG(x) cout<<"# # #     "<<x<<'\n';
#define fri(i,s,n) for(ll i=s;i<n;i++)
#define mod 1000000007LL
#define mod2 998244353
#define vct2 vector<vector<ll>>
#define vct vector<ll>
#define pll pair<ll,ll>
#define rz(n) resize(n)
#define all(v) v.begin(),v.end()
#define mkp make_pair
#define arrin(a,n) for(ll i = 0; i < n ; i++){cin>>a[i];}
#define arrout(a,n) for(ll i = 0;i < n; i++){cout<<a[i]<<" ";}
#define ex(s) cout<<s<<endl,exit(0);
#define inf 2e18
#define five 100005;
#define six 1000005;
#define prllclock cerr<<"Time "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
//_______________________________________________________________________________
ll f(ll r,ll k,ll c)
{
       ll rem = r-c;
       if(rem < 0){return 0;}
       return ((rem/k)+1);
}
int main()
{
       speed;
       ll t;
       cin>>t;
       while(t--)
       {
              ll a,b,q;
              cin>>a>>b>>q;
              if(a<b){swap(a,b);}
              ll answ[q];
              fri(it,0,q)
              {
                     ll l,r;
                     cin>>l>>r;
                     if((a%b) == 0){
                            cout<<0<<endl;
                     }
                     int i = 0;
                     ll ans = 0;
                     ll lcm = (a*b)/(__gcd(a,b));
                     while(i < b)
                     {
                            int k = 0;
                            while(1)
                            {
                                   if(k*b + i >= a){break;}
                                   ans += f(r,lcm,k*b+i) - f(l-1,lcm,k*b+i);
                                   k++;
                            }
                            i++;
                     }
                     answ[it] = (r-l+1)-ans;
              }
              fri(i,0,q){
                     cout<<answ[i]<<" ";
              }
              cout<<endl;
       }
}
