Problem: For given three points p1, p2, p, find the reflection point x of p onto p1p2.

<image>

Constraints

* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.

Input


xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1


In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.

Output

For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.

Examples

Input

0 0 2 0
3
-1 1
0 1
1 1


Output

-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000


Input

0 0 3 4
3
2 5
1 4
0 3


Output

4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000

Solution (in PYTHON3):

Solution:
#これの元のprojectionは通ったが、これはおかしくなる。
#数字自体は誤差の範囲内に収まった正解なので、-e16みたいな表示になるせいだと思う
#たぶんベクトルのノルムで割る操作が悪い
#連立方程式を解くべきか
L1=input().split( )
[xp1,yp1,xp2,yp2]=[int(L1[0]),int(L1[1]),int(L1[2]),int(L1[3])]
direction=[xp2-xp1,yp2-yp1]
norm_direc=[(xp2-xp1)/((xp2-xp1)**2+(yp2-yp1)**2)**(1/2),(yp2-yp1)/((xp2-xp1)**2+(yp2-yp1)**2)**(1/2)]
q=int(input())
L=[]
for i in range(q):
    L2=input().split( )
    L.append([int(L2[0]),int(L2[1])])

Point=[]
for i in range(q):
    Point.append([(L[i][0]-xp1),(L[i][1]-yp1)])

Vec=[]
for i in range(q):
    Vec.append([L[i][0]-xp1,L[i][1]-yp1])
Proj=[]
for i in range(q):
    inter=Vec[i][0]*norm_direc[0]+Vec[i][1]*norm_direc[1]
    Proj.append([norm_direc[0]*inter,norm_direc[1]*inter])
#以下射影用
#L2=[]

#for i in range(q):
#    L2.append([xp1+Proj[i][0],yp1+Proj[i][1]])
#for i in range(q):
#    print(*L2[i])

vert=[]
for i in range(q):
    vert.append([Vec[i][0]-Proj[i][0],Vec[i][1]-Proj[i][1]])
L3=[]
for i in range(q):
    a=L[i][0]-2*vert[i][0]
    b=L[i][1]-2*vert[i][1]
    if -0.00000001<a<0.00000001:
        a=0
    if -0.00000001<b<0.00000001:
        b=0
    L3.append([a,b])

for i in range(q):
    print(*L3[i])