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;
stack<char> k;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  long long int cnt = 0;
  for (long long int i = 0; i < s.length(); i++) {
    if (k.empty() || s[i] == '(' || s[i] == '[' || s[i] == '<' || s[i] == '{')
      k.push(s[i]);
    else {
      char ch = k.top();
      if (ch == '(' || ch == '[' || ch == '<' || ch == '{') {
        if ((ch != '(' && s[i] == ')') || (ch != '{' && s[i] == '}') ||
            (ch != '[' && s[i] == ']') || (ch != '<' && s[i] == '>'))
          cnt++;
        k.pop();
      }
    }
  }
  if (k.empty())
    cout << cnt << "\n";
  else
    cout << "Impossible"
         << "\n";
}