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;
bool isopen(char c) { return (c == '(' || c == '[' || c == '<' || c == '{'); }
bool differ(char op, char cl) {
  switch (op) {
    case '(':
      return cl != ')';
    case '[':
      return cl != ']';
    case '<':
      return cl != '>';
    case '{':
      return cl != '}';
  }
}
int main() {
  string str;
  cin >> str;
  stack<char> open;
  int ans = 0;
  for (int i = 0; str[i]; i++) {
    if (isopen(str[i]))
      open.push(str[i]);
    else {
      if (open.empty())
        return cout << "Impossible", 0;
      else if (differ(open.top(), str[i]))
        ans++;
      open.pop();
    }
  }
  if (!open.empty())
    cout << "Impossible";
  else
    cout << ans;
  return 0;
}