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;
int main() {
  stack<char> open;
  string str;
  cin >> str;
  bool flag = 0;
  int ans = 0;
  for (int i = 0; i < (int)str.size(); i++) {
    if (str[i] == '<' || str[i] == '(' || str[i] == '{' || str[i] == '[') {
      open.push(str[i]);
    } else {
      if (open.empty()) {
        flag = 1;
        break;
      }
      char now = open.top();
      open.pop();
      switch (str[i]) {
        case '>':
          if (now != '<') ans++;
          break;
        case '}':
          if (now != '{') ans++;
          break;
        case ']':
          if (now != '[') ans++;
          break;
        case ')':
          if (now != '(') ans++;
          break;
      }
    }
  }
  if (flag == 1 || open.size() != 0) {
    printf("Impossible\n");
  } else {
    printf("%d\n", ans);
  }
  return 0;
}
