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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define int long long
#define ll long long
#define ull unsigned long long
#define ff first
#define ss second
#define pb push_back
#define mk make_pair
#define ld long double
#define y1 asdfghjkl
#define y2 asdfghjkq
#define y3 asdfghjkw
#define ufo ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
using namespace __gnu_pbds;

typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;

mt19937_64 mt_rnd_64(chrono::steady_clock::now().time_since_epoch().count());
long long rnd(long long l, long long r){
    return (mt_rnd_64() % (r-l+1)) + l;
}

int q,t;
int a,b,l,r,ans,nsk,k,x;

main()
{
    ufo;
    cin>>t;
    for(int Q = 0;Q < t;Q++){
        cin>>a>>b>>q;
        if(a > b)swap(a,b);
        nsk = a*b/(__gcd(a,b));
//        cout<<nsk<<'\n';
        for(int T = 0;T < q;T++){
            cin>>l>>r;
            ans = r-l+1;
            if(nsk == b){
                cout<<0<<' ';
                continue;
            }
            k = r/nsk-(l-1)/nsk;
            x = r/nsk;
            x *= nsk;
            k *= b;
            ans -= k;
            if(x+b-1 > r)ans += x+b-1-r;
            x = l/nsk;
            x *= nsk;
            if(x+b-1 >= l)ans -= (x+b-1)-l+1;
            cout<<ans<<' ';
        }
        cout<<'\n';
    }
}
/**
1
1 10 1
1400 1489

1
6 9 1
89 91


1
6 9 5
89 91
56 99
14 88
1488 1900
7 7
1 19 39 206 0
**/
