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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define ll          long long
#define pb          push_back
#define mp 			make_pair
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define vii         vector<pii>
#define mi          map<ll,ll>
#define mii         map<pii,ll>
#define all(a)      (a).begin(),(a).end()
#define fi          first
#define rall(a)      (a).rbegin(),(a).rend()
#define si          second
#define sz(x)       (ll)x.size()
#define endl        '\n'
#define mod        1000000007
#define inf        1000000000000000000
#define iinf        -1000000000000000000
#define puts(n)		cout<<n;
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define mset(m,v) memset(m,v,sizeof(m))
#define tej ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
using namespace std;

int main ()
{
tej;
ll n;
cin>>n;
while(n--){
ll a,b,c;
cin>>a>>b>>c;
ll p=a,q=b;
a=min(p,q);
b=max(p,q);
ll o=a*b/__gcd(a,b);
rep(i,0,c){
ll x,y;
cin>>x>>y;
ll u=y/o;
ll t=max(x,b)/o;
if(y<b)cout<<0<<" ";
else {
   ll ans=y-max(x,b)+1-b*(u-t);
    cout<<ans<<" " ;
}}
cout<<endl;



}

return 0;}
