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 main() {
  string str;
  cin >> str;
  stack<char> s;
  map<char, char> type = {{'(', ')'}, {'{', '}'}, {'<', '>'}, {'[', ']'}};
  int ans = 0;
  for (auto it : str) {
    if (it == '{' || it == '(' || it == '<' || it == '[') {
      s.push(it);
    } else {
      if (s.empty()) {
        cout << "Impossible\n";
        return 0;
      } else {
        if (type[s.top()] != it) {
          ans += 1;
        }
        s.pop();
      }
    }
  }
  if (s.empty()) {
    cout << ans << "\n";
    return 0;
  }
  cout << "Impossible\n";
  return 0;
}