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;
stack<int> sk;
bool checkF(char ch) {
  if (ch == '[' || ch == '(' || ch == '{' || ch == '<') return true;
  return false;
}
bool checkM(char c1, char c2) {
  if (c1 == '<' && c2 == '>') return true;
  if (c1 == '(' && c2 == ')') return true;
  if (c1 == '[' && c2 == ']') return true;
  if (c1 == '{' && c2 == '}') return true;
  return false;
}
int main() {
  string s;
  cin >> s;
  int len = s.length();
  int ans = 0;
  for (int i = 0; i < len; i++) {
    if (checkF(s[i]))
      sk.push(s[i]);
    else {
      if (sk.empty()) {
        cout << "Impossible\n";
        return 0;
      } else {
        if (checkM(sk.top(), s[i]))
          ;
        else
          ans++;
        sk.pop();
      }
    }
  }
  if (sk.empty())
    cout << ans << endl;
  else
    cout << "Impossible\n";
}