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;
void __print(long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long a[1000007];
vector<long long> v;
void seive() {
  for (long long i = 0; i <= 1000000; i++) a[i] = 1;
  a[0] = a[1] = 0;
  for (long long i = 2; i <= 1000; i++) {
    if (a[i] == 1) {
      v.push_back(i);
      for (long long j = 2 * i; j <= 1000000; j += i) a[j] = 0;
    }
  }
  for (long long i = 1001; i <= 1000000; i++) {
    if (a[i] == 1) v.push_back(i);
  }
}
long long toint(const string &s) {
  stringstream ss;
  ss << s;
  long long x;
  ss >> x;
  return x;
}
bool opening(char c) {
  if (c == '(' || c == '{' || c == '<' || c == '[') return true;
  return false;
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  stack<char> st;
  long long n = s.length();
  if (!opening(s[0])) return cout << "Impossible", 0;
  long long open = 0, close = 0;
  for (long long i = 0; i < n; i++) {
    if (opening(s[i]))
      open++;
    else
      close++;
  }
  if (open != close) return cout << "Impossible", 0;
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (opening(s[i]))
      st.push(s[i]);
    else {
      if (!st.empty()) {
        if (s[i] == ')' && st.top() != '(')
          cnt++, st.pop();
        else if (s[i] == '}' && st.top() != '{')
          cnt++, st.pop();
        else if (s[i] == ']' && st.top() != '[')
          cnt++, st.pop();
        else if (s[i] == '>' && st.top() != '<')
          cnt++, st.pop();
        else
          st.pop();
      } else {
        cout << "Impossible";
        return 0;
      }
    }
  }
  cout << cnt;
  return 0;
}