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;
const long long mod = 1e9 + 7, siz = 3e5 + 5;
long long n, m, ans, a[siz];
bool is_close(char c) {
  if (c == '>' or c == ')' or c == '}' or c == ']') return true;
  return false;
}
char opp(char c) {
  if (c == '<') return '>';
  if (c == '(') return ')';
  if (c == '{') return '}';
  if (c == '[') return ']';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  stack<char> st;
  for (char c : s) {
    if (is_close(c)) {
      if (st.size() == 0) return cout << "Impossible", 0;
      if (c != opp(st.top())) ans++;
      st.pop();
    } else
      st.push(c);
  }
  if (st.size() != 0) return cout << "Impossible", 0;
  cout << ans << endl;
}
