Codeforces Round #331 (Div. 2) A. Wilbur and Swimming Pool （判断组成最大矩形）

A. Wilbur and Swimming Pool

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 werenot erased by Wilbur's friend.

Each of the following n lines contains two integersx_i andy_i ( - 1000 ≤ x_i, y_i ≤ 1000) —the coordinates of thei-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.

Examples

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.

题意：给你n个点，问四个点能组成的最大矩形，点必须是顶点

思路：直接判断就好

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 10000010#define LL long long#define ll __int64#define INF 0x7fffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headstruct s{int x,y;}a[MAXN];int main(){int n,i;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)scanf("%d%d",&a[i].x,&a[i].y);if(n==1){printf("-1\n");continue;}else if(n==2){if(a[0].x==a[1].x||a[0].y==a[1].y){printf("-1\n");continue;}else{printf("%d\n",abs(a[0].x-a[1].x)*abs(a[0].y-a[1].y));}}else{s aa,bb;aa.x=INF;aa.y=INF;bb.x=-INF;bb.y=-INF;for(i=0;i<n;i++){aa.x=min(aa.x,a[i].x);aa.y=min(aa.y,a[i].y);bb.x=max(bb.x,a[i].x);bb.y=max(bb.y,a[i].y);}if(aa.x==bb.x||aa.y==bb.y)printf("-1\n");elseprintf("%d\n",abs(aa.x-bb.x)*abs(aa.y-bb.y));}}return 0;}