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 equals(char ch, char c) {
  switch (ch) {
    case '{':
      return c == '}';
    case '[':
      return c == ']';
    case '(':
      return c == ')';
    case '<':
      return c == '>';
  }
}
void query(char ch, stack<char> &sta, int &ans) {
  if (!equals(sta.top(), ch)) ans++;
  sta.pop();
}
int main() {
  ios_base::sync_with_stdio(false);
  stack<char> sta;
  string s;
  cin >> s;
  int ans = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '[' || s[i] == '<' || s[i] == '{' || s[i] == '(') {
      sta.push(s[i]);
      continue;
    }
    if (sta.empty()) {
      ans = -1;
      break;
    }
    switch (s[i]) {
      case '}':
        query('}', sta, ans);
        continue;
      case '>':
        query('>', sta, ans);
        continue;
      case ')':
        query(')', sta, ans);
        continue;
      case ']':
        query(']', sta, ans);
        continue;
    }
  }
  if (!sta.empty()) ans = -1;
  if (ans == -1)
    cout << "Impossible";
  else
    cout << ans;
  cout << "\n";
  return 0;
}
