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 const maxn = 1e6 + 5;
char str[maxn];
int main() {
  stack<int> s;
  while (~scanf("%s", str)) {
    int ans = 0;
    int length = strlen(str);
    for (int i = 0; i < length; i++) {
      if (str[i] == '[')
        s.push(str[i]);
      else if (str[i] == '{')
        s.push(str[i]);
      else if (str[i] == '<')
        s.push(str[i]);
      else if (str[i] == '(')
        s.push(str[i]);
      else if (str[i] == ')') {
        if (!s.empty()) {
          if (s.top() != '(') ans++;
          s.pop();
        } else {
          ans = -1;
          break;
        }
      } else if (str[i] == ']') {
        if (!s.empty()) {
          if (s.top() != '[') ans++;
          s.pop();
        } else {
          ans = -1;
          break;
        }
      } else if (str[i] == '}') {
        if (!s.empty()) {
          if (s.top() != '{') ans++;
          s.pop();
        } else {
          ans = -1;
          break;
        }
      } else if (str[i] == '>') {
        if (!s.empty()) {
          if (s.top() != '<') ans++;
          s.pop();
        } else {
          ans = -1;
          break;
        }
      }
    }
    if (!s.empty()) ans = -1;
    if (ans == -1)
      cout << "Impossible" << endl;
    else
      cout << ans << endl;
  }
  return 0;
}