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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int res;
string s;
vector<int> opens;
map<char, char> m;
int main() {
  m.insert(make_pair('(', ')'));
  m.insert(make_pair('[', ']'));
  m.insert(make_pair('{', '}'));
  m.insert(make_pair('<', '>'));
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(' or s[i] == '[' or s[i] == '{' or s[i] == '<')
      opens.push_back(s[i]);
    else if (!opens.empty() and m[opens.back()] == s[i])
      opens.pop_back();
    else if (opens.empty()) {
      cout << "Impossible";
      return 0;
    } else {
      opens.pop_back();
      res++;
    }
  }
  if (!opens.empty()) {
    cout << "Impossible";
    return 0;
  }
  cout << res;
}