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;
stack<char> s;
char input[1000000 + 100];
int main() {
  scanf("%s", input);
  int left = 0;
  int right = 0;
  bool flag = true;
  int len = (int)strlen(input);
  int res = 0;
  for (int i = 0; i < len; i++) {
    if (input[i] == '<' || input[i] == '{' || input[i] == '[' ||
        input[i] == '(') {
      left++;
      s.push(input[i]);
    } else if (input[i] == '>') {
      right++;
      if (s.empty()) {
        flag = false;
        break;
      }
      if (s.top() == '<') {
        s.pop();
      } else {
        res++;
        s.pop();
      }
    } else if (input[i] == '}') {
      right++;
      if (s.empty()) {
        flag = false;
        break;
      }
      if (s.top() == '{') {
        s.pop();
      } else {
        res++;
        s.pop();
      }
    } else if (input[i] == ')') {
      right++;
      if (s.empty()) {
        flag = false;
        break;
      }
      if (s.top() == '(') {
        s.pop();
      } else {
        res++;
        s.pop();
      }
    } else if (input[i] == ']') {
      right++;
      if (s.empty()) {
        flag = false;
        break;
      }
      if (s.top() == '[') {
        s.pop();
      } else {
        res++;
        s.pop();
      }
    }
  }
  if (right != left) {
    flag = false;
  }
  if (flag) {
    cout << res << endl;
  } else {
    cout << "Impossible" << endl;
  }
  return 0;
}