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 is_opening(char c) {
  switch (c) {
    case '(':
    case '[':
    case '{':
    case '<':
      return true;
      break;
    case ')':
    case ']':
    case '}':
    case '>':
      return false;
      break;
  }
}
bool is_match_bracket(char c1, char c2) {
  if (c1 == '(' && c2 == ')') {
    return true;
  } else if (c1 == '[' && c2 == ']') {
    return true;
  } else if (c1 == '{' && c2 == '}') {
    return true;
  } else if (c1 == '<' && c2 == '>') {
    return true;
  }
  return false;
}
int main(void) {
  string str;
  cin >> str;
  size_t len = str.length();
  int ans = 0;
  stack<char> s;
  bool imp = false;
  for (int i = 0; i < len; ++i) {
    bool is_cur_opening = is_opening(str.at(i));
    if (!is_cur_opening) {
      if (!s.empty() && is_opening(s.top())) {
        if (!is_match_bracket(s.top(), str.at(i))) {
          ++ans;
        }
        s.pop();
      } else {
        imp = true;
        break;
      }
    } else {
      s.push(str.at(i));
    }
  }
  if (imp || !s.empty()) {
    cout << "Impossible" << endl;
  } else {
    cout << ans << endl;
  }
  return 0;
}