$(x2,Y2)$ is given by

$d=\sqrt[\phantom{2}]{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Three points in the Cartesian plan can form a triangle only if the sum of the distances between any two pairs of points is strictly greater than the distance between the third pair of points.

Write a $C$ function that takes as arguments three pairs of points and return $1$ if the three points do form a valid triangle and $0$ otherwise

For example:

Test

Input Result

printf$("\%$d$",$ istriangle $(3,2,-2,-3,2,3)$

$2$

$3$