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;
string s;
int cnt, ans;
stack<char> st;
bool open(char c) {
  if (c == '<' || c == '{' || c == '[' || c == '(') return 1;
  return 0;
}
bool close(char c) {
  if (c == '>' || c == '}' || c == ']' || c == ')') return 1;
  return 0;
}
bool Pair(char a, char b) {
  if (a == '(' && b == ')') return 1;
  if (a == '[' && b == ']') return 1;
  if (a == '{' && b == '}') return 1;
  if (a == '<' && b == '>') return 1;
  return 0;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin >> s;
  for (int i = 0; i < (int)s.size(); ++i) {
    if (open(s[i])) ++cnt;
    if (close(s[i])) --cnt;
    if (cnt < 0) {
      cout << "Impossible\n";
      return 0;
    }
  }
  if (cnt != 0) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 0; i < (int)s.size(); ++i) {
    if (open(s[i])) {
      st.push(s[i]);
    }
    if (close(s[i])) {
      if (!Pair(st.top(), s[i])) {
        ++ans;
      }
      st.pop();
    }
  }
  cout << ans;
  return 0;
}