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;
string s;
map<char, char> mp;
stack<char> sta;
long long ans;
int main() {
  mp['<'] = '>';
  mp['{'] = '}';
  mp['('] = ')';
  mp['['] = ']';
  cin >> s;
  for (int i = s.size() - 1; i >= 0; i--) {
    if (s[i] == ']' || s[i] == ')' || s[i] == '>' || s[i] == '}')
      sta.push(s[i]);
    else {
      if (sta.size() == 0) {
        cout << "Impossible";
        return 0;
      }
      if (mp[s[i]] != sta.top()) ans++;
      sta.pop();
    }
  }
  if (sta.size() != 0) {
    cout << "Impossible";
  } else
    cout << ans;
  return 0;
}
