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;
#define ll long long 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int main()
{
    IOS;
    ll tt;
    cin >> tt;
    while(tt--)
    {
        ll a,b,q;
        cin >> a >> b >> q;
        ll lcm=(a*b)/(__gcd(a,b));
        for(ll i=1;i<=q;i++)
        {
            ll l,r;
            cin >> l >> r;
            ll c=0;
            ll idx=l;
            while(idx<max(a,b))
            {
                ll o1=(idx%a)%b;
                ll o2=(idx%b)%a;
                if(o1==o2)
                {
                    c++;
                }
                idx++;
            }
            c+=((r/lcm)-((l-1)/lcm))*max(a,b);
            ll ans=0;
            ans+=(r-l+1)-c;
            // cout <<lcm << " " <<  c << endl;
            cout << max(ans,0ll) << " ";
        }
        cout << endl;
    }
    return 0;
}


