Codeforces Round #340 (Div. 2)D. Polyline

D. Polyline

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are three points marked on the coordinate plane. The goal is to make a simple polyline, without self-intersections and self-touches, such that it passes through all these points. Also, the polyline must consist of only segments parallel to the coordinate axes. You are to find the minimum number of segments this polyline may consist of.

Input

Each of the three lines of the input contains two integers. The i-th line contains integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th point. It is guaranteed that all points are distinct.

Output

Print a single number — the minimum possible number of segments of the polyline.

Examples

input

1 -11 11 2

output

1

input

-1 -1-1 34 3

output

2

input

1 12 33 2

output

3

Note

The variant of the polyline in the first sample:The variant of the polyline in the second sample:The variant of the polyline in the third sample:

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/* ***********************************************Author       : rycCreated Time : 2016-08-11 ThursdayFile Name    : E:\acm\codeforces\340D.cppLanguage     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#include<map>#include<stack>using namespace std;typedef long long LL;typedef pair<int,int> pii;const int maxn=10;struct Point{    int x,y;}A[maxn];bool cmp(Point a,Point b){    if(a.x==b.x)        return a.y<b.y;    return a.x<b.x;}bool cmp1(Point a,Point b){    if(a.y==b.y)        return a.x<b.x;    return a.y<b.y;}int main(){    for(int i=1;i<=3;++i){        scanf("%d%d",&A[i].x,&A[i].y);    }    if((A[1].x==A[2].x&&A[2].x==A[3].x)||(A[1].y==A[2].y&&A[1].y==A[3].y)){        printf("%d\n",1);    }    else if((A[1].x!=A[2].x&&A[1].x!=A[3].x&&A[2].x!=A[3].x)&&(A[1].y!=A[2].y&&A[1].y!=A[3].y&&A[2].y!=A[3].y)){        printf("%d\n",3);    }    else {        sort(A+1,A+4,cmp);        if(A[1].x==A[2].x&&(A[3].y>=A[2].y||A[3].y<=A[1].y)){            printf("%d\n",2);        }        else if(A[2].x==A[3].x&&(A[1].y<=A[2].y||A[1].y>=A[3].y)){            printf("%d\n",2);        }        else {            sort(A+1,A+4,cmp1);            if(A[1].y==A[2].y&&(A[3].x>=A[2].x||A[3].x<=A[1].x)){                printf("%d\n",2);            }            else if(A[2].y==A[3].y&&(A[1].x<=A[2].x||A[1].x>=A[3].x)){                printf("%d\n",2);            }            else {                printf("%d\n",3);            }        }    }    return 0;}