Problem: 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 <image>.

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long xxx1, yyy1, xxx2, yyy2, xxx3, yyy3;
double eps = 1e-9;
struct Point {
  long long x, y;
  Point(long long _x = 0, long long _y = 0) {
    x = _x;
    y = _y;
  }
};
Point operator+(Point A, Point B) { return Point(A.x + B.x, A.y + B.y); }
Point operator-(Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
Point operator*(Point A, double p) { return Point(p * A.x, p * A.y); }
Point operator/(Point A, double p) { return Point(A.x / p, A.y / p); }
Point operator-(Point A) { return Point(-A.x, -A.y); }
long long Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
long long getDis(long long xx1, long long yy1, long long xx2, long long yy2) {
  return (xx1 - xx2) * (xx1 - xx2) + (yy1 - yy2) * (yy1 - yy2);
}
int main() {
  while (~scanf("%I64d%I64d%I64d%I64d%I64d%I64d", &xxx1, &yyy1, &xxx2, &yyy2,
                &xxx3, &yyy3)) {
    long long d1 = getDis(xxx1, yyy1, xxx2, yyy2);
    long long d2 = getDis(xxx2, yyy2, xxx3, yyy3);
    Point v1 = Point(xxx2 - xxx1, yyy2 - yyy1);
    Point v2 = Point(xxx3 - xxx1, yyy3 - yyy1);
    long long judge = Cross(v1, v2);
    if (d1 == d2 && judge)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}