output_description: Print the number of seconds after which the hand of every clock point directly
  upward again.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      T_1
      :  
      T_N
attempt: def delete_head_zeros(n):
      n = str(n)
      l = len(n)
      if "." in n:
          l = n.find(".")
      head_zeros = 0
      for i in range(l - 1):
          if n[i] == "0":
              head_zeros += 1
          else:
              break
  
      return n[head_zeros:]
  
  
  # compare a, b
  # a, b: int or string
  def bigint_compare(a, b):
      a = delete_head_zeros(a)
      b = delete_head_zeros(b)
      if len(a) > len(b):
          return 1
      elif len(a) < len(b):
          return -1
      else:
          if a > b:
              return 1
          elif a < b:
              return -1
          else:
              return 0
  
  
  # calculate a + b
  # a, b: int or string
  def bigint_plus(a, b):
      a = str(a)
      b = str(b)
  
      d = max([len(a), len(b)])
      a = "0" * (d - len(a)) + a
      b = "0" * (d - len(b)) + b
  
      ans = ""
      carry = 0
      for i in range(d):
          s = int(a[-i - 1]) + int(b[-i - 1]) + carry
          carry = s // 10
          ans = str(s % 10) + ans
      else:
          if carry:
              ans = str(carry) + ans
  
      return ans
  
  
  # calculate a - b
  # a, b: int or string
  def bigint_minus(a, b):
      a = str(a)
      b = str(b)
      M = []
      m = []
      sign = ""
  
      if len(a) > len(b) or (len(a) == len(b) and a >= b):
          [M, m] = [a, b]
      else:
          [M, m] = [b, a]
          sign = "-"
      m = "0" * (len(M) - len(m)) + m
  
      ans = ""
      borrow = 0
      for i in range(len(M)):
          s = int(M[-i - 1]) - int(m[-i - 1]) - borrow
          if s < 0:
              borrow = 1
              s += 10
          else:
              borrow = 0
          ans = str(s) + ans
  
      return sign + delete_head_zeros(ans)
  
  
  # calculate a * b
  # a, b: int or string
  def bigint_multiply(a, b):
      a = str(a)
      b = str(b)
  
      md = []
      carry = 0
      for j in range(len(b)):
          mj = ""
          for i in range(len(a)):
              m = int(a[-i - 1]) * int(b[-j - 1]) + carry
              carry = m // 10
              mj = str(m % 10) + mj
          else:
              if carry:
                  mj = str(carry) + mj
          md.append(mj)
  
      ans = 0
      for k in range(len(md)):
          ans = bigint_plus(md[k] + "0" * k, ans)
  
      return ans
  
  
  # calculate a / b to d digits after decimal point
  # a, b, d: int or string
  def bigint_divide(a, b, d=0):
      a = str(a)
      b = str(b)
      d = int(d)
      if d < 0:
          d = 0
  
      ans = ""
      r = ""
      for i in range(len(a) + d):
          q = 0
          if i < len(a):
              r += a[i]
          elif i == len(a):
              ans += "."
              r += "0"
          else:
              r += "0"
  
          if bigint_compare(r, b) == -1:
              ans += str(q)
          else:
              while bigint_compare(r, b) >= 0:
                  r = bigint_minus(r, b)
                  q += 1
              ans += str(q)
  
      return delete_head_zeros(ans)
  
  
  def gcd(a, b):
      if bigint_compare(a, b) >= 0:
          M = a
          m = b
      else:
          M = b
          m = a
      if m == "0":
          return M
      else:
          q = bigint_divide(M, m)
          r = bigint_minus(M, bigint_multiply(m, q))
          return gcd(m, r)
  
  
  def lcm(a, b):
      return bigint_divide(bigint_multiply(a, b), gcd(a, b))
  
  
  def main():
      N = int(input())
      T = [input() for i in range(N)]
      ans = 1
      for i in range(len(T)):
          ans = lcm(T[i], ans)
  
      print(ans)
  
  
  main()
  
problem_description: Statement
  
  We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in
  exactly T_i seconds.  
  Initially, the hand of every clock stands still, pointing directly upward.  
  Now, Dolphin starts all the clocks simultaneously.  
  In how many seconds will the hand of every clock point directly upward again?
samples: [{"input": "2\n    2\n    3", "output": "6\n    \n\nWe have two clocks. The time when the hand of each clock points upward is as\nfollows:\n\n  * Clock 1: 2, 4, 6, ... seconds after the beginning\n  * Clock 2: 3, 6, 9, ... seconds after the beginning\n\nTherefore, it takes 6 seconds until the hands of both clocks point directly\nupward.\n\n* * *"}, {"input": "5\n    2\n    5\n    10\n    1000000000000000000\n    1000000000000000000", "output": "1000000000000000000"}]