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;
void solve() {
  string s;
  cin >> s;
  int n = s.length();
  s = ' ' + s;
  stack<char> st;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (st.empty() || s[i] == '{' || s[i] == '(' || s[i] == '[' ||
        s[i] == '<') {
      st.push(s[i]);
      continue;
    } else {
      char ch = st.top();
      if (ch == '{' || ch == '(' || ch == '[' || ch == '<') {
        if ((ch == '{' && s[i] == '}') || (ch == '(' && s[i] == ')') ||
            (ch == '[' && s[i] == ']') || (ch == '<' && s[i] == '>'))
          ;
        else
          ans++;
        st.pop();
      } else
        st.push(s[i]);
    }
  }
  if (!st.empty()) {
    puts("Impossible");
    return;
  }
  printf("%d", ans);
}
int main() {
  solve();
  return 0;
}
