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 std::abs;
using std::array;
using std::cerr;
using std::cin;
using std::cout;
using std::make_pair;
using std::map;
using std::max;
using std::min;
using std::pair;
using std::set;
using std::string;
using std::vector;
template <typename T>
T input() {
  T res;
  cin >> res;
  return res;
}
int get_type(char ch) {
  switch (ch) {
    case '<':
      return 1;
    case '(':
      return 2;
    case '[':
      return 3;
    case '{':
      return 4;
    case '>':
      return -1;
    case ')':
      return -2;
    case ']':
      return -3;
    case '}':
      return -4;
    default:
      return 0;
  }
}
int main() {
  std::iostream::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  string str = input<string>();
  uint32_t ans = 0;
  vector<int> stack;
  for (uint32_t i = 0; i != str.size(); ++i) {
    int tp = get_type(str[i]);
    assert(tp != 0);
    if (tp > 0)
      stack.push_back(tp);
    else {
      if (stack.empty()) {
        ans = UINT32_MAX;
        break;
      }
      if (stack.back() != -tp) ++ans;
      stack.pop_back();
    }
  }
  if (not stack.empty()) ans = UINT32_MAX;
  if (ans == UINT32_MAX)
    cout << "Impossible\n";
  else
    cout << ans << "\n";
  return 0;
}