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_open(char a) { return (a == '(' || a == '<' || a == '{' || a == '['); }
bool is_closed(char a) {
  return (a == ')' || a == '>' || a == '}' || a == ']');
}
int main() {
  string S;
  int n, i, answer = 0;
  list<char> L;
  cin >> S;
  n = S.length();
  bool m = true;
  for (i = 0; i < n; i++) {
    if (is_open(S[i]))
      L.push_back(S[i]);
    else if (is_closed(S[i])) {
      if (!L.empty()) {
        if (S[i] == ')') {
          if (L.back() != '(') ++answer;
          L.pop_back();
        } else if (S[i] == ']') {
          if (L.back() != '[') ++answer;
          L.pop_back();
        } else if (S[i] == '}') {
          if (L.back() != '{') ++answer;
          L.pop_back();
        } else if (S[i] == '>') {
          if (L.back() != '<') ++answer;
          L.pop_back();
        }
      } else {
        m = false;
        break;
      }
    }
  }
  if (!L.empty()) {
    m = false;
  }
  if (m)
    cout << answer << endl;
  else
    cout << "Impossible" << endl;
  return 0;
}
