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
s = input()
s1 = []

for i in s:
    if i == "(":
        s1.append(1)
    elif i == "[":
        s1.append(2)
    elif i == "{":
        s1.append(3)
    elif i == "<":
        s1.append(4)

    elif i == ")":
        s1.append(-1)
    elif i == "]":
        s1.append(-2)
    elif i == "}":
        s1.append(-3)
    else:
        s1.append(-4)


m = []
c =0
for char in s1:
    if len(m)>0:
        if m[-1]>0:
            if m[-1]==-char:
                m.pop()
            elif  char<0 :
                m.pop()
                c+=1
            else:
                m.append(char)
        else:
            m.append(char)
    else:
        m.append(char)

if len(m)>0:
    print("Impossible")
else:
    print(c)