output_description: 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.
input_description: 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.
attempt: # 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])
  
problem_description: Reflection
  
    
  
  For given three points p1, p2, p, find the reflection point x of p onto p1p2.
  
  ![](https://judgeapi.u-aizu.ac.jp/resources/images/IMAGE2_CGL_CGL_1_B)
samples: [{"input": "0 0 2 0\n    3\n    -1 1\n    0 1\n    1 1", "output": "-1.0000000000 -1.0000000000\n    0.0000000000 -1.0000000000\n    1.0000000000 -1.0000000000"}, {"input": "0 0 3 4\n    3\n    2 5\n    1 4\n    0 3", "output": "4.2400000000 3.3200000000\n    3.5600000000 2.0800000000\n    2.8800000000 0.8400000000"}]