Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = int(1e9);
const int MOD = int(1e9) + 7;
const long long N = 1000010;
const long long INFll = 1ll * INF * INF;
const long double ldINF = 1e+018;
const long double EPS = 0.000000001;
const int sdx[4] = {-1, 0, 1, 0};
const int sdy[4] = {0, 1, 0, -1};
const long double PI = 3.1415926535897932384;
template <typename T>
ostream& operator<<(ostream& out, pair<T, T>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, pair<T, T>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i] << endl;
    else
      out << a[i] << endl;
  return out;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  stack<char> st;
  set<char> open, close;
  open.insert('(');
  open.insert('[');
  open.insert('{');
  open.insert('<');
  close.insert(')');
  close.insert(']');
  close.insert('}');
  close.insert('>');
  map<char, char> m;
  m['('] = ')';
  m['['] = ']';
  m['<'] = '>';
  m['{'] = '}';
  m[')'] = '(';
  m[']'] = '[';
  m['>'] = '<';
  m['}'] = '{';
  int ans1 = 0;
  for (int i = 0; i < s.size(); i++)
    if (open.count(s[i]))
      st.push(s[i]);
    else {
      if (st.size() == 0) {
        ans1 = INF;
        break;
      }
      if (st.top() != m[s[i]]) ans1++;
      st.pop();
    }
  if (!st.empty()) ans1 = INF;
  int ans2 = 0;
  while (!st.empty()) st.pop();
  for (int i = s.size() - 1; i >= 0; i--)
    if (close.count(s[i]))
      st.push(s[i]);
    else {
      if (st.size() == 0) {
        ans2 = INF;
        break;
      }
      if (st.top() != m[s[i]]) ans2++;
      st.pop();
    }
  if (!st.empty()) ans2 = INF;
  if (min(ans1, ans2) == INF)
    cout << "Impossible\n";
  else
    cout << min(ans1, ans2) << endl;
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  return 0;
}