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 <numeric>
#define ll long long
#define MAX 100000
using namespace std;
/*ll power(ll x,ll y)
{
    ll t=1;
    for(ll i=1;i<=y;i++)
    {
        t*=x;
    }
    return t;
}*/
void solve()
{
    ll a,b,q,l,r,x,i,y;
    cin>>a>>b>>q;
    while(q--)
    {
        cin>>l>>r;
        if(r<max(a,b) || a==b)
        {
            cout<<0<<" ";
        }
        else
        {
            x=0;
            i=1;
            y=(a*b)/(__gcd(a,b));
            while(y*i<r)
            {
                if(y*i>=l)
                    x+=(min(max(a,b),r-(y*i)+1));
                i++;
            }
            cout<<min((r-max(a,b)+1),(r-l+1))-x<<" ";
        }
    }

    return;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
   ll t;
    cin>>t;
   while(t--)
    {
      solve();
      cout<<endl;
    }
    //solve();
    return 0;
}
