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;
int main() {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  map<char, bool> mp = {{'<', 1}, {'{', 1}, {'[', 1}, {'(', 1},
                        {']', 0}, {'}', 0}, {'>', 0}, {')', 0}};
  map<char, char> m = {{'<', '>'}, {'{', '}'}, {'[', ']'}, {'(', ')'}};
  stack<int> p;
  int ans = 0;
  for (auto c : s) {
    if (mp[c])
      p.push(c);
    else {
      if (p.empty()) {
        cout << "Impossible\n";
        return 0;
      } else if (c != m[p.top()])
        ans++;
      p.pop();
    }
  }
  if (p.empty())
    cout << ans << endl;
  else
    cout << "Impossible\n";
  return 0;
}