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>
bool openingBracket(char ch) {
  return ch == '<' || ch == '(' || ch == '[' || ch == '{';
}
int type(char ch) {
  if (ch == '<' || ch == '>') return 0;
  if (ch == '(' || ch == ')') return 1;
  if (ch == '[' || ch == ']') return 2;
  if (ch == '{' || ch == '}') return 3;
}
int main() {
  std::vector<char> s;
  int count = 0;
  for (char ch = getchar(); ch > ' '; ch = getchar()) {
    if (openingBracket(ch))
      s.push_back(ch);
    else {
      if (s.size() == 0) {
        printf("Impossible\n");
        return 0;
      }
      char prev = s.back();
      s.pop_back();
      if (type(ch) != type(prev)) count++;
    }
  }
  if (s.size() != 0)
    printf("Impossible\n");
  else
    printf("%d\n", count);
}