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
import java.util.*;

public class C {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    String s = scanner.next();
    int res = 0;
    Deque<Character> st = new ArrayDeque<>();
    for (int i = 0; i < s.length(); i++) {
      char chr = s.charAt(i);
      if (chr == '<' || chr == '[' || chr == '{' || chr == '(') {
        st.push(chr);
      } else {
        if (st.size() == 0) {
          System.out.println("Impossible");
          return;
        }
        char schr = st.pop();
        if (!((schr == '<' && chr == '>') || (schr == '[' && chr == ']') || (schr == '{' && chr == '}') || (schr == '(' && chr == ')'))) {
          res++;
        }
      }
    }
    if (st.size() != 0) {
      System.out.println("Impossible");
    } else {
      System.out.println(res);
    }
  }
}