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;
class STACK {
 private:
  struct node {
    char s;
    node *next;
  };
  node *top = NULL;

 public:
  char push(char s) {
    node *newest = new node;
    newest->s = s;
    if (top == NULL) {
      newest->next = NULL;
      top = newest;
    } else {
      node *aux = top->next;
      top->next = newest;
      top->next->next = aux;
      int aux2;
      aux2 = top->s;
      top->s = top->next->s;
      top->next->s = aux2;
    }
  }
  char pull() {
    if (top != NULL) top = top->next;
  }
  char pop() {
    if (top != NULL) return top->s;
  }
  int empty() {
    if (top == NULL)
      return 1;
    else
      return 0;
  }
};
int main() {
  STACK opening_brackets;
  STACK closing_brackets;
  STACK another_opening_brackets;
  STACK another_closing_brackets;
  string STRING;
  cin >> STRING;
  int i, c = 1, ob = 0, cb = 0, k = 0;
  for (i = 0; i < STRING.size(); i++) {
    if (STRING[i] == '<' || STRING[i] == '{' || STRING[i] == '[' ||
        STRING[i] == '(') {
      ob++;
      opening_brackets.push(STRING[i]);
    } else {
      cb++;
      closing_brackets.push(STRING[i]);
      if ((opening_brackets.pop() == '(' && closing_brackets.pop() == ')') ||
          (opening_brackets.pop() == '{' && closing_brackets.pop() == '}') ||
          (opening_brackets.pop() == '<' && closing_brackets.pop() == '>') ||
          (opening_brackets.pop() == '[' && closing_brackets.pop() == ']')) {
        opening_brackets.pull();
        closing_brackets.pull();
      } else if (!((opening_brackets.pop() == '(' &&
                    another_closing_brackets.pop() == ')') ||
                   (opening_brackets.pop() == '{' &&
                    another_closing_brackets.pop() == '}') ||
                   (opening_brackets.pop() == '<' &&
                    another_closing_brackets.pop() == '>') ||
                   (opening_brackets.pop() == '[' &&
                    another_closing_brackets.pop() == ']'))) {
        k++;
        opening_brackets.pull();
        closing_brackets.pull();
      }
    }
    if (cb > ob) c = 0;
  }
  if (c == 0 || STRING.size() % 2 > 0 || cb != ob)
    cout << "Impossible";
  else
    cout << k;
  return 0;
}