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;
string pairs = "{}[]()<>";
char getP(char c) {
  for (int i = 0; i < pairs.size(); i++) {
    if (pairs[i] == c) return pairs[i ^ 1];
  }
}
bool isOpen(char c) {
  for (int i = 0; i < pairs.size(); i++) {
    if (pairs[i] == c) return i % 2 == 0;
  }
}
stack<char> st;
int main() {
  string s;
  cin >> s;
  int res = 0;
  for (int i = 0; i < s.size(); i++) {
    if (isOpen(s[i])) {
      st.push(s[i]);
    } else {
      if (st.empty()) {
        puts("Impossible");
        return 0;
      }
      char c = st.top();
      st.pop();
      if (s[i] != getP(c)) {
        res++;
      }
    }
  }
  if (!st.empty()) {
    puts("Impossible");
  } else {
    printf("%d ", res);
  }
  return 0;
}
