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;
typedef long long ll;
const int maxx=1e6+7;
char s[maxx];
int main()
{
    ll t,a,b,q,l,r;
    cin>>t;
    while(t--)
    {
        cin>>a>>b>>q;
        while(q--)
        {
            cin>>l>>r;
            ll g=__gcd(a,b);
            ll temp=a*b/__gcd(a,b),m=max(a,b);
            for(int i=l;i<=r;i++)
                if(i%a%b==i%b%a) cout<<i<<endl;
            if(r<m||g==min(a,b)) cout<<0<<" ";
            else
            {
                ll ans=m*(r/temp)+min(m-1,r%temp)-m*((l-1)/temp)-min(m-1,(l-1)%temp);
                cout<<(r-l+1)-ans<<" ";
            }
        }
        cout<<endl;
    }
}