CodeForces 596A

Description

After making bad dives into swimming pools, Wilbur wants to build a swimming

pool in the shape of a rectangle in his backyard. He has set up coordinate axes,

and he wants the sides of the rectangle to be parallel to them. Of course, the

area of the rectangle must be positive. Wilbur had all four vertices of the planned

pool written on a paper, until his friend came along and erased some of the

vertices.Now Wilbur is wondering, if the remaining n vertices of the initial rectangle

give enough information to restore the area of the planned swimming pool.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 4) — the number of

vertices that were not erased by Wilbur's friend.

Each of the following n lines contains two integers x_i and y_i ( - 1000 ≤ x_i, y_i ≤ 1000)

—the coordinates of the i-th vertex that remains. Vertices are given in an arbitrary

order.It's guaranteed that these points are distinct vertices of some rectangle, that

has positive area and which sides are parallel to the coordinate axes.

Output

Print the area of the initial rectangle if it could be uniquely determined by the points

remaining. Otherwise, print  - 1.

Sample Input

Input

2
0 0
1 1

Output

1

Input

1
1 1

Output

-1

Hint

In the first sample, two opposite corners of the initial rectangle are given, and that

gives enough information to say that the rectangle is actually a unit square.

In the second sample there is only one vertex left and this is definitely not enough

to uniquely define the area.

一道签到题，然而我却wa了很多次，我一直以为是让我判断长方形，发现还要输出面积。。。。

 #include<iostream>
 #include<cmath>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 struct node
 {
     int x,y;
 }a[];
 int main()
 {
     int n,i,j;
     cin>>n;
     for(i=;i<=n;i++)
         cin>>a[i].x>>a[i].y;
     if(n==)
     {
         cout<<"-1"<<endl;
     }
     else if(n==)
     {
         if(a[].x==a[].x||a[].y==a[].y)
         {
             cout<<"-1"<<endl;
         }
         else
             cout<<abs(a[].x-a[].x)*abs(a[].y-a[].y)<<endl;
     }
     else
     {
         for(i=;i<=n;i++)
         {
             for(j=;j<=n;j++)
             {
                 if(a[i].x!=a[j].x&&a[i].y!=a[j].y)
                 {
                     cout<<abs(a[i].x-a[j].x)*abs(a[i].y-a[j].y)<<endl;
                     return  ;
                 }
             }
         }
         cout<<"-1"<<endl;
     }
     return ;
 }