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 gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int ar[500050];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc = 1;
  while (tc--) {
    string s;
    cin >> s;
    int n = s.size();
    if (s[0] == ']' or s[0] == '}' or s[0] == '>' or s[0] == ')')
      cout << "Impossible\n";
    else {
      int open = 0, close = 0;
      for (int i = 0; i < n; i++) {
        if (s[i] == ']' or s[i] == '}' or s[i] == '>' or s[i] == ')')
          close++;
        else
          open++;
      }
      if (open != close or n % 2) {
        cout << "Impossible\n";
        return 0;
      }
      stack<char> st;
      int cnt = 0;
      for (int i = 0; i < n; i++) {
        if (s[i] == '[' or s[i] == '{' or s[i] == '<' or s[i] == '(')
          st.push(s[i]);
        else {
          if (st.empty()) {
            cout << "Impossible\n";
            return 0;
          }
          if (st.top() == '(' and s[i] != ')')
            cnt++;
          else if (st.top() == '{' and s[i] != '}')
            cnt++;
          else if (st.top() == '[' and s[i] != ']')
            cnt++;
          else if (st.top() == '<' and s[i] != '>')
            cnt++;
          st.pop();
        }
      }
      cout << cnt << endl;
    }
  }
  return 0;
}