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;
bool brothers(char const &c1, char const &c2) {
  switch (c1) {
    case '[':
      return c2 == ']';
    case '{':
      return c2 == '}';
    case '(':
      return c2 == ')';
    case '<':
      return c2 == '>';
  }
}
bool is_open_bracket(char const &c1) {
  return (c1 == '(' || c1 == '[' || c1 == '{' || c1 == '<');
}
int main() {
  string input;
  cin >> input;
  int counter = 0;
  stack<char> c;
  for (int i = 0; i < input.length(); ++i) {
    if (is_open_bracket(input[i])) {
      c.push(input[i]);
    } else {
      if (c.empty()) {
        cout << "Impossible";
        return 0;
      }
      if (brothers(c.top(), input[i])) {
      } else {
        ++counter;
      }
      c.pop();
    }
  }
  if (c.empty()) {
    cout << counter;
  } else {
    cout << "Impossible";
  }
  return 0;
}