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;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, i, j, k;
  string st;
  cin >> st;
  long long ln = st.size(), pre = 0, post = 0, cnt = 0;
  stack<char> s;
  for (i = 0; i < ln; i++) {
    if (st[i] == '(' || st[i] == '{' || st[i] == '<' || st[i] == '[') {
      s.push(st[i]);
      pre++;
    } else {
      post++;
      if (pre > 0) {
        if (s.top() == '<' && st[i] != '>') {
          cnt++;
        } else if (s.top() == '(' && st[i] != ')') {
          cnt++;
        } else if (s.top() == '{' && st[i] != '}') {
          cnt++;
        } else if (s.top() == '[' && st[i] != ']') {
          cnt++;
        }
        s.pop();
        pre--;
        post--;
      }
    }
  }
  if (pre == 0 && post == 0)
    cout << cnt << endl;
  else
    cout << "Impossible" << endl;
  return 0;
}
