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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  stack<char> st;
  int cnt = 0;
  bool ok = true;
  for (auto c : s) {
    if (c == '(' || c == '<' || c == '{' || c == '[')
      st.push(c);
    else {
      if (st.empty()) {
        ok = false;
        break;
      } else if (st.top() == '(' && c == ')' || st.top() == '<' && c == '>' ||
                 st.top() == '{' && c == '}' || st.top() == '[' && c == ']')
        st.pop();
      else {
        cnt++;
        st.pop();
      }
    }
  }
  if (st.size()) ok = false;
  if (ok)
    cout << cnt;
  else
    cout << "Impossible";
  return 0;
}
