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;
bool is_o(char ch) {
  return (ch == '{' || ch == '[' || ch == '(' || ch == '<');
}
bool diff(char a, char b) {
  if (a == '{') return b != '}';
  if (a == '[') return b != ']';
  if (a == '(') return b != ')';
  if (a == '<') return b != '>';
}
int main() {
  string s;
  cin >> s;
  int res = 0, op = 0;
  int i = 0;
  stack<char> ss;
  while (i < s.length()) {
    while (is_o(s[i])) {
      ss.push(s[i]);
      op++;
      i++;
    }
    while (!is_o(s[i]) && i < s.length()) {
      op--;
      if (ss.empty()) {
        cout << "Impossible";
        return 0;
      }
      if (diff(ss.top(), s[i])) res++;
      ss.pop();
      i++;
    }
  }
  if (op) {
    cout << "Impossible";
    return 0;
  }
  cout << res;
  return 0;
}
