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 main(int argc, char **argv) {
  vector<char> v;
  char ca_open[4] = {40, 60, 91, 123};
  char ca_close[4] = {41, 62, 93, 125};
  int i_chnge = 0;
  int i_cntr = 0;
  string s;
  cin >> s;
  if ((s[0] == ca_close[0]) || (s[0] == ca_close[1]) || (s[0] == ca_close[2]) ||
      (s[0] == ca_close[3])) {
    cout << "Impossible" << endl;
    return 0;
  } else {
    if (s.size() % 2 != 0) {
      cout << "Impossible" << endl;
      return 0;
    }
    for (int i = 0; i < s.size(); ++i) {
      ((s[i] == ca_open[0]) || (s[i] == ca_open[1]) || (s[i] == ca_open[2]) ||
       (s[i] == ca_open[3]))
          ? ++i_cntr
          : --i_cntr;
      if (i_cntr < 0) {
        cout << "Impossible" << endl;
        return 0;
      }
      if ((s[i] == ca_close[0]) || (s[i] == ca_close[1]) ||
          (s[i] == ca_close[2]) || (s[i] == ca_close[3])) {
        if (v.size() > 0 && abs(s[i] - v[v.size() - 1]) > 2) ++i_chnge;
        v.pop_back();
      } else
        v.push_back(s[i]);
    }
    if (i_cntr != 0) {
      cout << "Impossible" << endl;
      return 0;
    }
  }
  cout << i_chnge << endl;
  return 0;
}