Problem: You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.

The following definition of a regular bracket sequence is well-known, so you can be familiar with it.

Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.

For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string s RBS.

Input

The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.

Output

If it's impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Examples

Input

[&lt;}){}


Output

2

Input

{()}[]


Output

0

Input

]]


Output

Impossible

Solution (in PYTHON3):

Solution:
s = list(input())
def isopen(c):
  return c == "<" or c == "{" or c == "[" or c == "(";
def match(a, b):
  return a == "<" and b == ">" or a == "{" and b == "}" or a == "[" and b == "]" or a == "(" and b == ")"
def isRbs():
  br = []
  if isopen(s[0]) == False:
    print("Impossible")
    return False
  count = 0
  for c in s:
    open = isopen(c)
    if open == True:
      br.append(c)
    elif len(br) > 0 and open == False:
      b = br.pop()
      if match(b, c) == False:
        count += 1
  if len(br) == 0:
    print(count)
  else:
    print("Impossible")

if len(s) % 2 != 0:
  print("Impossible")
else:
  isRbs()