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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Solution {
  static boolean isOpening(Character c) {
    return c == '(' || c == '[' || c == '{' || c == '<';
  }
  
  static boolean isAngle(Character c) {
    return c == '<' || c == '>';
  }
  
  static boolean isCurly(Character c) {
    return c == '{' || c == '}';
  }
  
  static boolean isSquare(Character c) {
    return c == '[' || c == ']';
  }
  
  static boolean isRound(Character c) {
    return c == '(' || c == ')';
  }
  
  static boolean sameKind(Character c1, Character c2) {
    return (isCurly(c1) && isCurly(c2)) ||
        (isAngle(c1) && isAngle(c2)) ||
        (isSquare(c1) && isSquare(c2)) ||
        (isRound(c1) && isRound(c2));
  }
  
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String text = sc.next();

    int result = 0;
    Stack<Character> stack = new Stack<Character>();
    for (Character c : text.toCharArray()) {
      if (isOpening(c)) {
        stack.push(c);
      } else if (!stack.empty()) {
        if (!sameKind(stack.peek(), c)) {
          ++result;
        }
        stack.pop();
      } else {
        result = -1;
        break;
      }
    }

    if (result == -1 || !stack.empty()) {
      System.out.println("Impossible");
    } else {
      System.out.println(result);
    }
  }
}