B. Eastern Exhibition- Codeforces Round ＃703 (Div. 2)

Problem - 1486B - Codeforces

B. Eastern Exhibition

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You and your friends live in nn houses. Each house is located on a 2D plane, in a point with integer coordinates. There might be different houses located in the same point. The mayor of the city is asking you for places for the building of the Eastern exhibition. You have to find the number of places (points with integer coordinates), so that the summary distance from all the houses to the exhibition is minimal. The exhibition can be built in the same point as some house. The distance between two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) is |x1−x2|+|y1−y2||x1−x2|+|y1−y2|, where |x||x| is the absolute value of xx.

Input

First line contains a single integer tt (1≤t≤1000)(1≤t≤1000) — the number of test cases.

The first line of each test case contains a single integer nn (1≤n≤1000)(1≤n≤1000). Next nn lines describe the positions of the houses (xi,yi)(xi,yi) (0≤xi,yi≤109)(0≤xi,yi≤109).

It's guaranteed that the sum of all nn does not exceed 10001000.

Output

For each test case output a single integer - the number of different positions for the exhibition. The exhibition can be built in the same point as some house.

Example

input

Copy

6
3
0 0
2 0
1 2
4
1 0
0 2
2 3
3 1
4
0 0
0 1
1 0
1 1
2
0 0
1 1
2
0 0
2 0
2
0 0
0 0

output

Copy

1
4
4
4
3
1

Note

Here are the images for the example test cases. Blue dots stand for the houses, green — possible positions for the exhibition.

First test case.

Second test case.

Third test case.

Fourth test case.

Fifth test case.

Sixth test case. Here both houses are located at (0,0)(0,0).

about X,Y The choice of the two does not affect each other , Manhattan is just XY The sum of the differences , So just let X,Y Take the minimum respectively , It turns into a one-dimensional problem , On a line segment , The smallest point of the sum of odd points Manhattan distance is at its midpoint , The minimum sum of Manhattan distances of even points is within the closed interval formed by the middle two points .

# include<iostream>
# include<complex.h>
# include<string.h>
# include<cstring>
# include<vector>
# include<algorithm>
# include<iomanip>

using namespace std;
typedef complex<double>cp;
# define mod 9999991
typedef long long int  ll;


ll x[1010],y[1010];

int main ()
{


    int t;

    cin>>t;

    while(t--)
    {
        int n;

        cin>>n;

        for(int i=1;i<=n;i++)
        {
            cin>>x[i];

            cin>>y[i];

        }

        if(n&1)
        {
            cout<<1<<'\n';

            continue;

        }
        sort(x+1,x+1+n);

        sort(y+1,y+1+n);


        ll L,R;

        L=x[n/2+1]-x[n/2]+1;

        R=y[n/2+1]-y[n/2]+1;

        cout<<L*R<<'\n';


    }

    return 0;

}