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
rr=raw_input
rrI = lambda: int(rr())
rrM = lambda: map(int,rr().split())
debug=0
if debug:
    fi = open('t.txt','r')
    rr=lambda: fi.readline().replace('\n','')

left = set()
pair = {}
for i,j in zip('<[{(','>]})'):
    pair[i] = j
    pair[j] = i
    left.add(i)
    
def solve(S):
    stack = []
    ans = 0
    while S:
        e = S.pop()
        if e in left:
            stack.append(e)
        else:
            if not stack: return "Impossible"
            f = stack.pop()
            ans += pair[e]!=f
    if stack: return "Impossible"
    return ans



S = list(rr())[::-1]
print solve(S)
