output_description: Print the number of pairs with integer product A_i \cdot A_j (and i < j).
  
  * * *
input_description: Input is given from Standard Input in the following format.
  
      
      
      N
      A_1
      A_2
      \vdots
      A_N
attempt: cnt = 100
  res = []
  
  
  def one():
      global cnt
      global res
      res.append(f"< 2 0 {cnt}")
      cnt += 1
      return cnt - 1
  
  
  def twice(x):
      global cnt
      global res
      res.append(f"+ {x} {x} {cnt}")
      cnt += 1
      return cnt - 1
  
  
  def plus(x, y):
      global cnt
      global res
      res.append(f"+ {x} {y} {cnt}")
      cnt += 1
      return cnt - 1
  
  
  def gt(x, y):
      global cnt
      global res
      res.append(f"< {x} {y} {cnt}")
      cnt += 1
      return cnt - 1
  
  
  def tmp():
      global cnt
      cnt += 1
      return cnt - 1
  
  
  def mult(x, y):
      global one
      global cnt
      global res
      res.append(f"< {x} {one} {cnt}")
      res.append(f"< {cnt} {y} {cnt + 1}")
      cnt += 2
      return cnt - 1
  
  
  A = 0
  B = 1
  zero = tmp()
  one = one()
  two = twice(one)
  four = twice(two)
  eight = twice(four)
  A4 = gt(eight, plus(A, one))
  A3 = gt(plus(four, twice(twice(twice(A4)))), plus(A, one))
  A2 = gt(plus(two, plus(twice(twice(twice(A4))), twice(twice(A3)))), plus(A, one))
  A1 = gt(
      plus(one, plus(plus(twice(twice(twice(A4))), twice(twice(A3))), twice(A2))),
      plus(A, one),
  )
  
  B4 = gt(eight, plus(B, one))
  B3 = gt(plus(four, twice(twice(twice(B4)))), plus(B, one))
  B2 = gt(plus(two, plus(twice(twice(twice(B4))), twice(twice(B3)))), plus(B, one))
  B1 = gt(
      plus(one, plus(plus(twice(twice(twice(B4))), twice(twice(B3))), twice(B2))),
      plus(B, one),
  )
  
  A1B1 = mult(one, mult(A1, B1))
  A1B2 = mult(one, mult(A1, B2))
  A1B3 = mult(one, mult(A1, B3))
  A1B4 = mult(one, mult(A1, B4))
  
  A2B1 = mult(one, mult(A2, B1))
  A2B2 = mult(one, mult(A2, B2))
  A2B3 = mult(one, mult(A2, B3))
  A2B4 = mult(one, mult(A2, B4))
  
  A3B1 = mult(A3, B1)
  A3B2 = mult(A3, B2)
  A3B3 = mult(A3, B3)
  A3B4 = mult(A3, B4)
  
  A4B1 = mult(A4, B1)
  A4B2 = mult(A4, B2)
  A4B3 = mult(A4, B3)
  A4B4 = mult(A4, B4)
  
  AB1 = A1B1
  AB2 = plus(A1B2, A2B1)
  AB3 = plus(plus(A1B3, A2B2), A3B1)
  AB4 = plus(plus(A1B4, A2B3), plus(A3B2, A4B1))
  AB5 = plus(plus(A2B4, A3B3), A4B2)
  AB6 = plus(A3B4, A4B3)
  AB7 = A4B4
  
  t1 = AB1
  t2 = twice(AB2)
  t3 = twice(twice(AB3))
  t4 = twice(twice(twice(AB4)))
  t5 = twice(twice(twice(twice(AB5))))
  t6 = twice(twice(twice(twice(twice(AB6)))))
  t7 = twice(twice(twice(twice(twice(twice(AB7))))))
  
  ans = plus(plus(plus(t1, t2), plus(t3, t4)), plus(plus(t5, t6), t7))
  res.append(f"+ 2 {ans}  2")
  
  print(len(res))
  for i in res:
      print(i)
  
problem_description: Statement
  
  You are given N real values A_1, A_2, \ldots, A_N. Compute the number of pairs
  of indices (i, j) such that i < j and the product A_i \cdot A_j is integer.
samples: [{"input": "5\n    7.5\n    2.4\n    17.000000001\n    17\n    16.000000000", "output": "3\n    \n\nThere are 3 pairs with integer product:\n\n  * 7.5 \\cdot 2.4 = 18\n  * 7.5 \\cdot 16 = 120\n  * 17 \\cdot 16 = 272\n\n* * *"}, {"input": "11\n    0.9\n    1\n    1\n    1.25\n    2.30000\n    5\n    70\n    0.000000001\n    9999.999999999\n    0.999999999\n    1.000000001", "output": "8"}]