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;
stack<char> s;
string str;
map<char, int> mp1, mp2;
int main() {
  int i, j;
  int cnt = 0;
  int flag = 0;
  mp1['['] = mp2[']'] = 0;
  mp1['('] = mp2[')'] = 1;
  mp1['{'] = mp2['}'] = 2;
  mp1['<'] = mp2['>'] = 3;
  cin >> str;
  for (i = 0; i < str.size(); i++) {
    if (str[i] == '[' || str[i] == '(' || str[i] == '{' || str[i] == '<') {
      s.push(str[i]);
    } else {
      if (s.empty()) {
        flag = 1;
        break;
      } else {
        if (mp1[s.top()] != mp2[str[i]]) cnt++;
        s.pop();
      }
    }
  }
  if (!s.empty()) flag = 1;
  if (flag)
    printf("Impossible\n");
  else
    printf("%d\n", cnt);
  return 0;
}