getc str fff ica esc closed com describes cal

You are given n points with integer coordinates on the plane. Points are given in a way such that there is no triangle, formed by any three of these n points, which area exceeds S.

Alyona tried to construct a triangle with integer coordinates, which contains all n points and which area doesn‘t exceed 4S

, but, by obvious reason, had no success in that. Please help Alyona construct such triangle. Please note that vertices of resulting triangle are not necessarily chosen from n given points.

Input

In the first line of the input two integers n and S (3?≤?n?≤?5000, 1?≤?S?≤?10¹⁸) are given — the number of points given and the upper bound value of any triangle‘s area, formed by any three of given n

points.

The next n lines describes given points: i^th of them consists of two integers x_i and y_i (?-?10⁸?≤?x_i,?y_i?≤?10⁸) — coordinates of i^th point.

It is guaranteed that there is at least one triple of points not lying on the same line.

Output

Print the coordinates of three points — vertices of a triangle which contains all n

points and which area doesn‘t exceed 4S.

Coordinates of every triangle‘s vertex should be printed on a separate line, every coordinate pair should be separated by a single space. Coordinates should be an integers not exceeding 10⁹ by absolute value.

It is guaranteed that there is at least one desired triangle. If there is more than one answer, print any of them.

Example

Input

4 1
0 0
1 0
0 1
1 1

Output

-1 0
2 0
0 2

Note

要找個三角形把所有點都包在裏面

畫畫圖就會發現，只要在這些點中找個面積最大的三角形，然後對三條邊都翻折過去變成一個4倍大的三角形就好了

面積最大的三角形找法有點迷……一開始就選123號點，然後每次嘗試把一個點替換之後面積會不會變大

一直做到不再變大為止。

這玩意似乎有什麽定理 可以證明2-stable point一定屬於3-stable point?

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<deque>
 8 #include<set>
 9 #include<map>
10 #include<ctime>
11 #define LL long long
12 #define inf 0x7ffffff
13 #define pa pair<int,int>
14 #define pi 3.1415926535897932384626433832795028841971
15 using namespace std;
16 inline LL read()
17 {
18     LL x=0,f=1;char ch=getchar();
19     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
20     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
21     return x*f;
22 }
23 inline void write(LL a)
24 {
25     if (a<0){printf("-");a=-a;}
26     if (a>=10)write(a/10);
27     putchar(a%10+‘0‘);
28 }
29 inline void writeln(LL a){write(a);printf("\n");}
30 LL n,k;
31 struct point{
32     LL x,y;
33 }p[100010];
34 int a,b,c;
35 point operator +(point a,point b){return (point){a.x+b.x,a.y+b.y};}
36 point operator -(point a,point b){return (point){a.x-b.x,a.y-b.y};}
37 inline LL calc(int a,int b,int c)
38 {
39     LL x1=p[c].x-p[a].x,y1=p[c].y-p[a].y,x2=p[b].x-p[a].x,y2=p[b].y-p[a].y;
40     LL ans=x1*y2-x2*y1;
41     if (ans<0)ans=-ans;
42     return ans;
43 }
44 int main()
45 {
46     n=read();k=read();
47     for (int i=1;i<=n;i++)p[i].x=read(),p[i].y=read();
48     a=1;b=2;c=3;
49     bool refresh=0;
50     while (!refresh)
51     {
52         refresh=1;
53         for (int i=1;i<=n;i++)
54         if (i!=a&&i!=b&&i!=c)
55         {
56             if (calc(a,b,c)<calc(i,b,c))a=i,refresh=0;
57             else if (calc(a,b,c)<calc(a,i,c))b=i,refresh=0;
58             else if (calc(a,b,c)<calc(a,b,i))c=i,refresh=0;
59         }
60     }
61     point d=p[a]+p[b]-p[c],e=p[a]+p[c]-p[b],f=p[c]+p[b]-p[a];
62     printf("%lld %lld\n",d.x,d.y);
63     printf("%lld %lld\n",e.x,e.y);
64     printf("%lld %lld\n",f.x,f.y);
65 }

cf 682E

cf682E Alyona and Triangles