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;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char c) { return '\'' + string(1, c) + '\''; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) res += ", ";
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (unsigned i = 0; i < v.size(); i++) {
    if (!first) res += ", ";
    first = false;
    res += to_string(bool(v[i]));
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
string to_str(const int &n) {
  ostringstream stm;
  stm << n;
  return stm.str();
}
string to_str(const long long &n) {
  ostringstream stm;
  stm << n;
  return stm.str();
}
template <class T>
string tostring(T x, int len = 0) {
  stringstream ss;
  ss << x;
  string r = ss.str();
  if (int((r).size()) < len) r = string(len - int((r).size()), '0') + r;
  return r;
}
template <class T>
void convert(string x, T &r) {
  stringstream ss(x);
  ss >> r;
}
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long modinv(long long a) { return powmod(a, 1000000007 - 2); }
long long modinv(long long a, long long m) { return powmod(a, m - 2); }
long long nCkMp(long long n, long long k) {
  long long numerator = 1, denominator = 1;
  for (int i = 0; i < k; i++) numerator = (numerator * (n - i)) % 1000000007;
  for (int i = 1; i <= k; i++) denominator = (denominator * i) % 1000000007;
  return (numerator * modinv(denominator)) % 1000000007;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
int TT = 1, tt;
clock_t t1;
void solve() {
  t1 = clock();
  string s;
  int res = 0;
  char tp;
  (cin >> (s));
  stack<char> q;
  for (auto i = (0); i < (int((s).size())); i++) {
    if (s[i] == '{' || s[i] == '[' || s[i] == '<' || s[i] == '(')
      q.push(s[i]), 404;
    else {
      404;
      if (q.empty()) {
        (cout << ("Impossible") << endl);
        return;
      }
      tp = q.top();
      if (tp == '}' || tp == ']' || tp == '>' || tp == ')') {
        (cout << ("Impossible") << endl);
        return;
      }
      q.pop();
      if (tp == '{' && s[i] != '}')
        res++;
      else if (tp == '[' && s[i] != ']')
        res++;
      else if (tp == '<' && s[i] != '>')
        res++;
      else if (tp == '(' && s[i] != ')')
        res++;
      404;
    }
  }
  if (q.empty())
    (cout << (res) << endl);
  else
    (cout << ("Impossible") << endl);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  for (tt = 0; tt < TT; tt++) solve();
  return EXIT_SUCCESS;
}