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 min(int a, int b) { return (a < b) ? a : b; }
int max(int a, int b) { return (a > b) ? a : b; }
int main() {
  int k = 0;
  char tmp;
  stack<char> c;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    tmp = s[i];
    if ((tmp == '{') || (tmp == '(') || (tmp == '[') || (tmp == '<')) {
      c.push(tmp);
      continue;
    }
    if (c.empty() == true) {
      cout << "Impossible";
      return 0;
    }
    if (tmp == '}') {
      if (c.top() != '{') k++;
      c.pop();
    }
    if (tmp == ')') {
      if (c.top() != '(') k++;
      c.pop();
    }
    if (tmp == ']') {
      if (c.top() != '[') k++;
      c.pop();
    }
    if (tmp == '>') {
      if (c.top() != '<') k++;
      c.pop();
    }
  }
  if (c.empty() == false)
    cout << "Impossible";
  else
    cout << k;
  return 0;
}
