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>
int main() {
  std::string s;
  std::cin >> s;
  std::stack<char> st;
  bool ok = true;
  int ans = 0;
  for (int i = 0; ok && i < s.length(); ++i) {
    switch (s[i]) {
      case '(':
      case '<':
      case '{':
      case '[':
        st.push(s[i]);
        break;
      case ')':
        if (st.empty()) {
          ok = false;
        } else {
          if (st.top() != '(') ans++;
          st.pop();
        }
        break;
      case '>':
        if (st.empty()) {
          ok = false;
        } else {
          if (st.top() != '<') ans++;
          st.pop();
        }
        break;
      case '}':
        if (st.empty()) {
          ok = false;
        } else {
          if (st.top() != '{') ans++;
          st.pop();
        }
        break;
      case ']':
        if (st.empty()) {
          ok = false;
        } else {
          if (st.top() != '[') ans++;
          st.pop();
        }
        break;
    }
  }
  if (st.empty() && ok) {
    std::cout << ans << '\n';
  } else {
    std::cout << "Impossible\n";
  }
  return 0;
}