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;
stack<char> s;
int main() {
  int open = 0, close = 0, ans = 0;
  char c;
  while (cin >> c) {
    if (c == '<' || c == '[' || c == '(' || c == '{' || s.empty()) {
      s.push(c);
      open++;
    } else if (s.top() == '<' || s.top() == '(' || s.top() == '[' ||
               s.top() == '{') {
      if (s.top() == '<' && c == '>' || s.top() == '[' && c == ']' ||
          s.top() == '(' && c == ')' || s.top() == '{' && c == '}')
        s.pop();
      else {
        ans++;
        s.pop();
      }
    }
  }
  if (!s.empty())
    cout << "Impossible";
  else
    cout << ans;
  return 0;
}
