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
st = input()

c = 0
b = r = s = l = 0
for i in st:
    if i in [ '[' , '<' , '{' , '(' ]:
        c += 1
    else:
        c -= 1
        
        if c < 0:
          break
        
ans = 0
if c != 0:
    print('Impossible')
else:
    stack = []
    for i in st:
        if i in [ '[' , '<' , '{' , '(' ]:
            stack.append(i)
        else:
            if stack[-1] == '(' and i == ')':
                stack.pop()
                continue
            elif stack[-1] == '<' and i == '>':
                stack.pop()
                continue
            elif stack[-1] == '{' and i == '}':
                stack.pop()
                continue
            elif stack[-1] == '[' and i == ']':
                stack.pop()
                continue
            else:
                stack.pop()
                ans += 1
    print(ans)