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
#include <bits/stdc++.h>
using namespace std;
int func(const string& str) {
  stack<char> s;
  int ret = 0;
  for (int i = 0; i < str.size(); ++i) {
    char ch = str[i];
    if (ch == '<' || ch == '{' || ch == '[' || ch == '(')
      s.push(ch);
    else {
      if (s.empty()) return -1;
      char top = s.top();
      s.pop();
      if (top == '<' && ch == '>')
        ;
      else if (top == '{' && ch == '}')
        ;
      else if (top == '[' && ch == ']')
        ;
      else if (top == '(' && ch == ')')
        ;
      else
        ++ret;
    }
  }
  if (s.size()) return -1;
  return ret;
}
int main() {
  string str;
  cin >> str;
  int sol = func(str);
  if (sol == -1)
    cout << "Impossible" << endl;
  else
    cout << sol << endl;
}
