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;
bool open(char s) {
  if (s == '<' || s == '{' || s == '(' || s == '[') return true;
  return false;
}
bool eq(char s1, char s2) {
  if (s1 == '<') {
    if (s2 == '>')
      return true;
    else
      return false;
  } else if (s1 == '{') {
    if (s2 == '}')
      return true;
    else
      return false;
  } else if (s1 == '[') {
    if (s2 == ']')
      return true;
    else
      return false;
  } else {
    if (s2 == ')')
      return true;
    else
      return false;
  }
}
int main() {
  string s;
  cin >> s;
  stack<char> so;
  int ans = 0;
  for (int i = 0; i < s.size(); i++) {
    if (open(s[i])) {
      so.push(s[i]);
    } else {
      if (so.empty()) {
        cout << "Impossible";
        return 0;
      }
      if (!eq(so.top(), s[i])) {
        ++ans;
      }
      so.pop();
    }
  }
  if (!so.empty()) {
    cout << "Impossible";
    return 0;
  }
  cout << ans;
  return 0;
}
