Score : $300$ points

Problem Statement

Consider a two-dimensional coordinate plane, where the $x$-axis is oriented to the right, and the $y$-axis is oriented upward.

In this plane, there is a quadrilateral without self-intersection. The coordinates of the four vertices are $(A_x,A_y)$, $(B_x,B_y)$, $(C_x,C_y)$, and $(D_x,D_y)$, in counter-clockwise order.

Determine whether this quadrilateral is convex.

Here, a quadrilateral is convex if and only if all four interior angles are less than $180$ degrees.

Constraints

• $-100 \leq A_x,A_y,B_x,B_y,C_x,C_y,D_x,D_y \leq 100$
• All values in input are integers.
• The given four points are the four vertices of a quadrilateral in counter-clockwise order.
• The quadrilateral formed by the given four points has no self-intersection and is non-degenerate. That is,- no two vertices are at the same coordinates;
  • no three vertices are colinear; and
  • no two edges that are not adjacent have a common point.
• no two vertices are at the same coordinates;
• no three vertices are colinear; and
• no two edges that are not adjacent have a common point.

Input

Input is given from Standard Input in the following format:

$A_x$ $A_y$

$B_x$ $B_y$

$C_x$ $C_y$

$D_x$ $D_y$

Output

If the given quadrilateral is convex, print Yes; otherwise, print No.

0 0
1 0
1 1
0 1

Yes

The given quadrilateral is a square, whose four interior angles are all $90$ degrees. Thus, this quadrilateral is convex.

0 0
1 1
-1 0
1 -1

No

The angle $A$ is $270$ degrees. Thus, this quadrilateral is not convex.