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;
char s[1000100];
stack<char> S;
int main() {
  gets(s);
  int t = 0, d;
  d = strlen(s);
  for (int i = 0; i < d; i++) {
    if (d % 2 != 0) {
      printf("Impossible\n");
      return 0;
    }
    if (s[i] == ']') {
      if (S.size() == 0) {
        printf("Impossible\n");
        return 0;
      }
      if (S.top() == '[')
        S.pop();
      else {
        t++;
        S.pop();
      }
    } else if (s[i] == ')') {
      if (S.size() == 0) {
        printf("Impossible\n");
        return 0;
      }
      if (S.top() == '(')
        S.pop();
      else {
        t++;
        S.pop();
      }
    } else if (s[i] == '>') {
      if (S.size() == 0) {
        printf("Impossible\n");
        return 0;
      }
      if (S.top() == '<')
        S.pop();
      else {
        t++;
        S.pop();
      }
    } else if (s[i] == '}') {
      if (S.size() == 0) {
        printf("Impossible\n");
        return 0;
      }
      if (S.top() == '{')
        S.pop();
      else {
        t++;
        S.pop();
      }
    } else
      S.push(s[i]);
  }
  if (S.size() != 0)
    printf("Impossible\n");
  else
    printf("%d\n", t);
  return 0;
}