NOKIA - Connecting Soldiers (Code Chef Easy Question)

Hello Everyone !!,
Welcome Back to our blog for another question and solution from codechef.
Today's Question is Connecting Soldiers , code named NOKIA.

The prerequisites for solving the question are simple logic , recursion .



PROBLEM

To protect people from evil, a long and tall wall was constructed a few years ago. But just a wall is not safe, there should also be soldiers on it, always keeping vigil. The wall is very long and connects the left and the right towers. There are exactly N spots (numbered 1 to N) on the wall for soldiers. The K^th spot is K miles far from the left tower and (N+1-K) miles from the right tower.

Given a permutation of spots P of {1, 2, ..., N}, soldiers occupy the N spots in that order. The P[i]^th spot is occupied before the P[i+1]^th spot. When a soldier occupies a spot, he is connected to his nearest soldier already placed to his left. If there is no soldier to his left, he is connected to the left tower. The same is the case with right side. A connection between two spots requires a wire of length equal to the distance between the two.

The realm has already purchased a wire of M miles long from Nokia, possibly the wire will be cut into smaller length wires. As we can observe, the total length of the used wire depends on the permutation of the spots P. Help the realm in minimizing the length of the unused wire. If there is not enough wire, output -1.

Input

First line contains an integer T (number of test cases, 1 ≤ T ≤ 10 ). Each of the next T lines contains two integers N M, as explained in the problem statement (1 ≤ N ≤ 30 , 1 ≤ M ≤ 1000).

Output

For each test case, output the minimum length of the unused wire, or -1 if the the wire is not sufficient.

Example

Input:
4
3 8
3 9
2 4
5 25

Output:
0
0
-1
5

Explanation:
In the 1st case, for example, the permutation P = {2, 1, 3} will use the exact 8 miles wires in total.

In the 2nd case, for example, the permutation P = {1, 3, 2} will use the exact 9 miles wires in total.

To understand the first two cases, you can see the following figures:






In the 3rd case, the minimum length of wire required is 5, for any of the permutations {1,2} or {2,1}, so length 4 is not sufficient.

In the 4th case, for the permutation {1, 2, 3, 4, 5} we need the maximum length of the wire = 20. So minimum possible unused wire length = 25 - 20 = 5.














Now , We look at the solution .
The solution is easy , we just need to find the maximum and minimum wire we can use in our permutations .
So the maximum case is  easier , just insert the soldiers at each end and continue inserting them alternately from each end .
By simple induction it can be proved , this will give the maximum length.

For the minimum length , we need to use a recursion as each soldier will be inserted at the middle position available then and then in the middle of the two segments formed and so on until all soldiers are inserted . This will guarantee the minimum length of wire.
If the given wire length lies between our bounds then we have an answer , otherwise it is not possible to achieve an answer and we output -1.


The following C++ code achieves this :-

#include <iostream>
using namespace std;
int func(int start,int end)
{
int mid=(start+end)/2;
if(start==end || end-start==1)
return 0;
else
return func(start,mid)+func(mid,end)+end-mid+mid-start;
}

int main() {
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        int min=func(0,n+1);
        int max=0;
        max=(n*n+3*n)/2;
        
        if(m>=min && m<=max)
        {
            cout<<0<<endl;
        }
        else if(m<min)
        {
            cout<<-1<<endl;
        }
        else
        {
            cout<<m-max<<endl;
        }
    }
// your code goes here
return 0;

}

Hope you liked it .
Do comment if you have any doubts and also follow and subscribe to my blog for much interesting updates and learning a lot of more stuff .