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>
#include <ext/pb_ds/detail/standard_policies.hpp>

using namespace std;
using namespace __gnu_pbds;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define pb push_back
#define ll long long
#define pii pair< int, int >
#define pli pair< ll, int >
#define fastc ios_base::sync_with_stdio(0)
int t, q;
ll a, b, l, r;
ll k, s, e, gcd;
ll res = 0;
ll cal( ll x, ll y){
    ll temp = x / k;
    return max(0LL, (y - max( x, temp*k + b) + 1LL));

}
int main(){
    cin >> t;
    while( t-- ){
        cin >> a >> b>> q;
        gcd = __gcd(a, b);
        k = (a/gcd)*b;
        while( q-- ){
            cin >> l >> r;
            res = 0;
            s = (l / k );
            e = (r / k );
            if( s != e ){
                res += (e - s - 1LL) * ( k - b);
                res += cal( l, ((s+1)*k - 1)) + cal( e*k , r);
            }else{ res += cal( l, r);}
            cout << res <<" ";
        }
        cout << "\n";
    }
    return 0;
}
