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 isopen(char c) { return (c == '{' || c == '[' || c == '<' || c == '('); }
char mirror(char c) {
  if (c == '}') return '{';
  if (c == ']') return '[';
  if (c == '>') return '<';
  if (c == ')') return '(';
  if (c == '{') return '}';
  if (c == '[') return ']';
  if (c == '<') return '>';
  if (c == '(') return ')';
}
char ch[1000001];
int ans;
string s;
int main() {
  cin >> s;
  int k = 0;
  if (s.length() % 2) return cout << "Impossible", 0;
  for (int i = 0; i < s.length(); i++) {
    if (isopen(s[i])) {
      k++;
      ch[k] = s[i];
    } else {
      k--;
      if (k < 0) {
        return cout << "Impossible", 0;
      }
      if (ch[k + 1] != mirror(s[i])) ans++;
    }
  }
  if (k != 0) return cout << "Impossible", 0;
  cout << ans;
  return 0;
}
