p02306 Incircle of a Triangle
Write a program which prints the central coordinate ($cx$,$cy$) and the radius $r$ of a incircle of a triangle which is constructed by three points ($x_1$, $y_1$), ($x_2$, $y_2$) and ($x_3$, $y_3$) on the plane surface.

Constraints

* $-10000 \leq x_i, y_i \leq 10000$
* The three points are not on the same straight line

Input

The input is given in the following format


$x_1$ $y_1$
$x_2$ $y_2$
$x_3$ $y_3$


All the input are integers.

Output

Print $cx$, $cy$ and $r$ separated by a single space in a line. The output values should be in a decimal fraction with an error less than 0.000001.

Examples

Input

1 -2
3 2
-2 0


Output

0.53907943898209422325 -0.26437392711448356856 1.18845545916395465278


Input

0 3
4 0
0 0


Output

1.00000000000000000000 1.00000000000000000000 1.00000000000000000000
```python
import math

x1, y1 = map(int, input().split())
x2, y2 = map(int, input().split())
x3, y3 = map(int, input().split())

#3辺の長さ
a = math.sqrt(((x1-x2)**2)+((y1-y2)**2))
b = math.sqrt(((x3-x2)**2)+((y3-y2)**2))
c = math.sqrt(((x1-x3)**2)+((y1-y3)**2))
s = (a+b+c)/2

r = math.sqrt(s*(s-a)*(s-b)*(s-c))/s #ヘロンの公式
xa, ya = x1+(x2-x1)*c/(b+c), y1+(y2-y1)*c/(b+c) #二等分線とaの交点
xb, yb = x2+(x3-x2)*a/(a+c), y2+(y3-y2)*a/(a+c)

if (xa-x3) != 0 and (xb-x1) != 0:
    a1, b1 = (ya-y3)/(xa-x3), -((ya-y3)/(xa-x3))*x3+y3 #二等分線の方程式
    a2, b2 = (yb-y1)/(xb-x1), -((yb-y1)/(xb-x1))*x1+y1
    cx = (b2-b1)/(a1-a2)
    cy = a1 * cx + b1

if xa-x3 == 0:
    cx = x3
    a2, b2 = (yb-y1)/(xb-x1), -((yb-y1)/(xb-x1))*x1+y1
    cy = a2 * cx + b2

if xb-x1 == 0:
    cx = x1
    a1, b1 = (ya-y3)/(xa-x3), -((ya-y3)/(xa-x3))*x3+y3
    cy = a1 * cx + b1

print(cx, cy, r)

```