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;
int getInt() {
  int x;
  scanf("%d", &x);
  return x;
}
long long getLongLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
double getDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
char getChar() {
  char x;
  scanf("%c", &x);
  return x;
}
template <typename t>
t abs(t a) {
  if (a >= 0) return a;
  return -a;
}
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream second(s);
  string x;
  while (getline(second, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << "  ";
  err(++it, args...);
}
void FastIO() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
char s[2000000];
int main() {
  int len = strlen(gets(s));
  int ans = 0, yes = 1;
  stack<char> stk;
  char db[256];
  db['<'] = '>';
  db['('] = ')';
  db['{'] = '}';
  db['['] = ']';
  db[']'] = '[';
  db['>'] = '<';
  db[')'] = '(';
  db['}'] = '{';
  for (int i = 0; i < len; i++) {
    if (s[i] == '(' || s[i] == '{' || s[i] == '[' || s[i] == '<')
      stk.push(s[i]);
    else {
      if (stk.empty()) {
        yes = 0;
        break;
      }
      if (stk.top() != db[s[i]]) ans++;
      stk.pop();
    }
  }
  if (yes && stk.empty())
    printf("%d\n", ans);
  else
    puts("Impossible");
  return 0;
}