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;
void solve() {
  string s;
  cin >> s;
  vector<char> v;
  long long k = 0;
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == '(' || s[i] == '[' || s[i] == '{' || s[i] == '<') {
      v.push_back(s[i]);
    } else {
      if (v.empty()) {
        cout << "Impossible";
        return;
      }
      if (s[i] == ')' && v.back() != '(') {
        k++;
      }
      if (s[i] == '}' && v.back() != '{') {
        k++;
      }
      if (s[i] == ']' && v.back() != '[') {
        k++;
      }
      if (s[i] == '>' && v.back() != '<') {
        k++;
      }
      v.pop_back();
    }
  }
  if (v.size()) {
    cout << "Impossible";
    return;
  }
  cout << k;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}