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;
map<char, char> m;
bool is(char ch) {
  if (ch == '<' || ch == '{' || ch == '(' || ch == '[') return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  m['>'] = '<';
  m[')'] = '(';
  m['}'] = '{';
  m[']'] = '[';
  stack<char> st;
  string a;
  cin >> a;
  int cnt = 0;
  for (int i = 0; i < a.size(); i++) {
    bool isopen = false;
    if (is(a[i])) isopen = true;
    if (st.empty()) {
      if (isopen)
        st.push(a[i]);
      else {
        cout << "Impossible" << endl;
        return 0;
      }
    } else {
      if (isopen)
        st.push(a[i]);
      else {
        char ch = st.top();
        st.pop();
        if (ch != m[a[i]]) cnt++;
      }
    }
  }
  if (st.size() > 0)
    cout << "Impossible" << endl;
  else
    cout << cnt;
}