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;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
template <typename A, typename B>
ostream &operator<<(ostream &cout, map<A, B> const &v) {
  for (auto x : v) cout << x.first << ": " << x.second << "\n";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<array<A, 2>> const &v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i][0] << " " << v[i][1] << "\n";
}
int32_t main() {
  string s;
  cin >> s;
  stack<char> ss;
  long long ans = 0;
  for (auto x : s) {
    if (x == '(' || x == '{' || x == '[' || x == '<')
      ss.push(x);
    else {
      if (ss.empty()) {
        cout << "Impossible";
        return 0;
      }
      if (ss.top() == '(' && x == ')')
        ss.pop();
      else if (ss.top() == '[' && x == ']')
        ss.pop();
      else if (ss.top() == '{' && x == '}')
        ss.pop();
      else if (ss.top() == '<' && x == '>')
        ss.pop();
      else if (ss.top() != '(' && x == ')') {
        ss.pop();
        ans++;
      } else if (ss.top() != '[' && x == ']') {
        ss.pop();
        ans++;
      } else if (ss.top() != '{' && x == '}') {
        ss.pop();
        ans++;
      } else if (ss.top() != '<' && x == '>') {
        ss.pop();
        ans++;
      }
    }
  }
  if (!ss.empty())
    cout << "Impossible";
  else
    cout << ans;
}