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;
stack<char> ss;
bool samechar(char a, char b) {
  if (a == '[' && b == ']') return true;
  if (a == '<' && b == '>') return true;
  if (a == '{' && b == '}') return true;
  if (a == '(' && b == ')') return true;
  return false;
}
int main() {
  string str;
  cin >> str;
  int ans = 0;
  int n = str.length();
  for (int i = 0; i < n; i++) {
    if (str[i] == '(' || str[i] == '[' || str[i] == '<' || str[i] == '{') {
      ss.push(str[i]);
      continue;
    }
    if (ss.size() == 0) {
      cout << "Impossible" << endl;
      return 0;
    }
    char z = ss.top();
    if (!samechar(z, str[i])) {
      ans++;
    }
    ss.pop();
  }
  if (ss.size() != 0) {
    cout << "Impossible" << endl;
    return 0;
  }
  cout << ans << endl;
}
