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 isOpen(char c) {
  if (c == '(' || c == '{' || c == '<' || c == '[') return true;
  return false;
}
bool isMatch(char o, char c) {
  if ((o == '(' && c == ')') || (o == '{' && c == '}') ||
      (o == '<' && c == '>') || (o == '[' && c == ']'))
    return true;
  return false;
}
int main() {
  string s;
  cin >> s;
  stack<char> stk;
  int cnt = 0;
  for (int i = 0; i < s.size(); i++) {
    if (isOpen(s[i]))
      stk.push(s[i]);
    else {
      if (stk.empty()) {
        cout << "Impossible";
        return 0;
      } else if (isMatch(stk.top(), s[i]))
        stk.pop();
      else {
        cnt++;
        stk.pop();
      }
    }
  }
  if (!stk.empty())
    cout << "Impossible";
  else
    cout << cnt;
}
