Rama is in love with geometry. So once he was playing with circles and rectangles. Given the center of circle and radius and also the co-ordinates of vertices of rectangle, he wants to check whether the rectangle lies inside the circle or not.

Note: If all the vertices are lie the circumference of circle then it should be considered to be inside.

Input:

First line will contain a positive integer t - number of test cases.
For each test case there will be 5 lines - first will contain three space separated integers radius r, x co-ordinate of center xc and y co-ordinate of circle yc . For next 4 lines there will be 2 space separated numbers each, first denoting the x co-ordinate of that vertex(Xi) and the y co-ordinate of that vertex(Yi) of the rectangle.

Output:

For each test case, output a single line containing “Yes” (without quotes) if the rectangle lies inside the circle, or “No” (without quotes) if it does not lie inside.

Constraints:

1 ≤ t ≤ 10^4
1 ≤ r ≤ 10^4
-10^4 ≤ xc , yc , Xi , Yi ≤ 10^4

SAMPLE INPUT
1
5 0 0
-3 2
2 2
2 -2
-3 -2

SAMPLE OUTPUT
Yes
t = int(raw_input())
for i in range(t):
	r,a,b = map(int,raw_input().split(" "))
	p,q = map(int,raw_input().split(" "))
	l,m = map(int,raw_input().split(" "))
	e,f = map(int,raw_input().split(" "))
	g,h = map(int,raw_input().split(" "))
	c = r*r
	if ((p-a)*(p-a) + (q-b)*(q-b)) > c  :
		print "No"
		continue
	if (l-a)*(l-a) + (m-b)*(m-b) > c  :
		print "No"
		continue
	if (e-a)*(e-a) + (f-b)*(f-b) > c  :
		print "No"
		continue
	if (g-a)*(g-a) + (h-b)*(h-b) > c  :
		print "No"
		continue
	print "Yes"