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[500];
stack<char> q;
int main() {
  cin >> s;
  bool flag = 0;
  int t = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '{' || s[i] == '[' || s[i] == '<' || s[i] == '(')
      t++;
    else
      t--;
    if (t < 0) flag = 1;
    cnt[s[i]]++;
  }
  int sum = 0;
  sum = cnt['{'] + cnt['['] + cnt['('] + cnt['<'];
  if (s.size() - sum != sum || flag) {
    cout << "Impossible";
  } else {
    int ans = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '{' || s[i] == '[' || s[i] == '<' || s[i] == '(')
        q.push(s[i]);
      else {
        if (q.empty()) {
          cout << "Impossible";
          return 0;
        }
        if (q.top() == '{' && s[i] == '}')
          q.pop();
        else if (q.top() == '(' && s[i] == ')')
          q.pop();
        else if (q.top() == '[' && s[i] == ']')
          q.pop();
        else if (q.top() == '<' && s[i] == '>')
          q.pop();
        else
          q.pop(), ans++;
      }
    }
    cout << ans;
  }
}