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;
vector<char> stek;
int main() {
  string str;
  bool solving = true;
  cin >> str;
  int xx;
  int oper = 0;
  int opened = 0;
  int closed = 0;
  int x = 0;
  for (int i = 0; i < str.length(); i++) {
    if (str[i] == '{' || str[i] == '[' || str[i] == '<' || str[i] == '(') {
      x++;
    } else {
      x--;
    }
    if (x < 0) {
      solving = false;
    }
  }
  if (x != 0) {
    solving = false;
  }
  if (!solving) {
    cout << "Impossible" << endl;
    return 0;
  } else {
    for (int i = 0; i < str.length(); i++) {
      if (str[i] == '{' || str[i] == '[' || str[i] == '<' || str[i] == '(') {
        if (str[i] == '{') stek.push_back(1);
        if (str[i] == '[') stek.push_back(2);
        if (str[i] == '<') stek.push_back(3);
        if (str[i] == '(') stek.push_back(4);
      } else if (str[i] == '}' || str[i] == ']' || str[i] == '>' ||
                 str[i] == ')') {
        if (str[i] == '}') xx = 1;
        if (str[i] == ']') xx = 2;
        if (str[i] == '>') xx = 3;
        if (str[i] == ')') xx = 4;
        int temp = stek.back();
        stek.pop_back();
        if (temp != xx) {
          oper++;
        }
      }
    }
    cout << oper << endl;
  }
  return 0;
}