Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool isLeft(char &c) { return c == '{' || c == '[' || c == '<' || c == '('; }
bool isOneSide(char &a, char &b) {
  return isLeft(a) && isLeft(b) || !isLeft(a) && !isLeft(b);
}
bool isOneType(char &a, char &b) {
  return (a == '{' && b == '}') || (a == '(' && b == ')') ||
         (a == '[' && b == ']') || (a == '<' && b == '>');
}
int main() {
  string str;
  cin >> str;
  stack<char> st, buff;
  for (long long i = 0; i < str.length(); i++) st.push(str[i]);
  char temp;
  long long count = 0;
  while (!st.empty()) {
    temp = st.top();
    st.pop();
    if (!isLeft(temp)) {
      buff.push(temp);
    } else {
      if (buff.empty()) {
        count = -1;
        break;
      }
      if (!isOneType(temp, buff.top())) {
        count++;
      }
      buff.pop();
    }
  }
  if (!buff.empty()) count = -1;
  cout << '\n';
  if (count == -1)
    cout << "Impossible";
  else
    cout << count;
  return 0;
}