Three points of known parallelogram for fourth point

Long time ago Alex created an interesting problem about parallelogram. The input data for this problem contained four integer points on the Cartesian plane, which defined the set of vertices of Some non-degenerate (positive area) parallelogram. Points not necessary were given in the order of clockwise or counterclockwise traversal.

Alex had very nice test for this problem, but are somehow happened that the last line of the the input were lost and now he had Only three off of four points of the original parallelogram. He remembers that test is so good this he asks you to the restore it given only these three points.

Input

The input consists of three lines, each containing a pair of integer coordinates x_{i and Yi (?-? 1000?≤? Xi,? Yi? ≤?1000). It's guaranteed that these three points does not lie on the same line and no both of the them coincide.}

Output

First print integer k -the number of ways to add one new integer point such that the obtained set defines Some parallelogram of positive area. There is no requirement for the points to being arranged in any special order (like traversal), they just define the set of V Ertices.

Then print k -lines, each containing a pair of integer-possible coordinates of the fourth point.

Example

Input

0 0
1 0
0 1

Output

3
1-1
-1 1
1 1
Code

#Include<stdio.h>#include <stdlib.h >int main  () {int x1,y1,x2,y2,x3,y3; while (scanf ( "%d%d%d%d%d%d", &X1,&Y1,&X2,&Y2,&X3,&Y3)!=eof) {printf (  "3\n"); printf ( "%d%d\n", x1+x2-x3,y1+y2-y3); printf ( "%d%d\n", x2+x3-x1,y2+y3-y1); printf ( "%d%d\n", x1+x3-x2,y1+y3-y2);} return 0;}         

Three points of known parallelogram for fourth point