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() {
  string s;
  cin >> s;
  string st, req = "{<[(";
  int ans = 0;
  for (auto i : s) {
    if (req.find(i) != string::npos) {
      st.push_back(i);
    } else {
      if (st.length()) {
        char c = st[st.length() - 1];
        st.pop_back();
        if (req.find(c) == string::npos) {
          printf("Impossible\n");
          return 0;
        }
        if (c == '<' && i != '>') ans++;
        if (c == '(' && i != ')') ans++;
        if (c == '[' && i != ']') ans++;
        if (c == '{' && i != '}') ans++;
      } else {
        printf("Impossible\n");
        return 0;
      }
    }
  }
  if (st.length() == 0)
    printf("%d\n", ans);
  else {
    printf("Impossible\n");
  }
}
