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 match(char a, char b) {
  if (a == '[' && b == ']')
    return 1;
  else if (a == '{' && b == '}')
    return 1;
  else if (a == '(' && b == ')')
    return 1;
  else if (a == '<' && b == '>')
    return 1;
  else
    return 0;
}
int main() {
  string str;
  int count = 0, n = 0;
  stack<char> s;
  cin >> str;
  int len = (int)str.size();
  for (int i = 0; i < len; i++) {
    if (str[i] == '<' || str[i] == '{' || str[i] == '(' || str[i] == '[')
      s.push(str[i]);
    else {
      if (s.empty()) {
        n++;
        break;
      } else {
        char c = s.top();
        int k = match(c, str[i]);
        if (k == 1)
          s.pop();
        else {
          count++;
          s.pop();
        }
      }
    }
  }
  int p = s.empty();
  if (n > 0 || p == 0)
    cout << "Impossible";
  else
    cout << count;
}
