codeforces 851B Arpa and an exam about geometry

B. Arpa and an exam about geometry

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Arpa is taking a geometry exam. Here is the last problem of the exam.

You are given three points a, b, c.

Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.

Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.

Input

The only line contains six integers ax, ay, bx, by, cx, cy (|ax|, |ay|, |bx|, |by|, |cx|, |cy| ≤ 109). It's guaranteed that the points are distinct.

Output

Print "Yes" if the problem has a solution, "No" otherwise.

You can print each letter in any case (upper or lower).

Examples

input

0 1 1 1 1 0

output

Yes

input

1 1 0 0 1000 1000

output

No

Note

In the first sample test, rotate the page around (0.5, 0.5) by .

In the second sample test, you can't find any solution.

题意：给定a，b，c三个点，求是否存在一个点d使a绕d旋转一定角度与b重合，b绕d旋转相同角度与c重合

因为a，b，c的位置使固定的且a，b，c不重合，要满足这个条件判断a，b，c不在一条直线上且Lab=Lbc（可以画图证明），因为数据较大，计算长度相等是相乘可能爆int

#pragma comment(linker,"/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<queue>#include<deque>#include<set>#include<map>#include<cmath>#include<vector>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> PII;#define pi acos(-1.0)#define eps 1e-10#define pf printf#define sf scanf#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r#define ess tree[rt]#define _s second#define _f first#define all(x) (x).begin,(x).end#define mem(i,a) memset(i,a,sizeof i)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define mi ((l+r)>>1)#define sqr(x) ((x)*(x))const int inf=0x3f3f3f3f;ll a,b,c,d,e,f;ll length(ll a,ll b,ll c,ll d){    return sqr(a-c)+sqr(b-d);//不用返回double，计算平方和是否相等即可}bool check(ll a,ll b,ll c,ll d,ll e,ll f){    if((d-b)*(e-c)==(f-d)*(c-a))//计算斜率相等用乘法，防止除数为0        return 0;    return 1;}int main(){    while(~sf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&e,&f))    {        if(length(a,b,c,d)==length(c,d,e,f)&&check(a,b,c,d,e,f))            puts("Yes");        else            puts("No");    }    return 0;}