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 ats;
string S;
vector<char> br;
int main() {
  cin >> S;
  for (int i = 0; i < S.length(); i++) {
    if (S[i] == '<' || S[i] == '{' || S[i] == '[' || S[i] == '(') {
      br.push_back(S[i]);
    } else if (br.empty()) {
      cout << "Impossible" << endl;
      return 0;
    } else if ((S[i] == '>' && br.back() != '<') ||
               (S[i] == '}' && br.back() != '{') ||
               (S[i] == ']' && br.back() != '[') ||
               (S[i] == ')' && br.back() != '(')) {
      ats++;
      br.pop_back();
    } else {
      br.pop_back();
    }
  }
  if (br.empty())
    printf("%d\n", ats);
  else
    cout << "Impossible" << endl;
}
