【Codeforces Round 331 (Div 2)A】【水题】Wilbur and Swimming Pool 给出矩形1~4个顶点让你还原矩形

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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 test(s)
input
20 01 1
output
1
input
11 1
output
-1

Note

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.

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=0,M=0,Z=1e9+7,ms63=1061109567;int x[4];int y[4];int n;int main(){while(~scanf("%d",&n)){int maxx=-1e9;int minx=1e9;int maxy=-1e9;int miny=1e9;for(int i=1;i<=n;++i){scanf("%d%d",&x[i],&y[i]);gmax(maxx,x[i]);gmax(maxy,y[i]);gmin(minx,x[i]);gmin(miny,y[i]);}int ans=(maxx-minx)*(maxy-miny);if(ans==0)ans=-1;printf("%d\n",ans);}return 0;}/*【trick&&吐槽】我用%d输出LL，竟然还是AC了！好感动！也好不小心啊！【题意】我们有一个矩形。现在告诉你这个矩形的n（1<=n<=4）个点，问你，我们能否还原这个矩形。【类型】水题【分析】你还打算讨论？噗！其实我们只要看看我们能否找出矩形的左下角和右上角， 求出其面积就好啦！*/

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