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
#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b;
  char s[1000006];
  scanf("%1000000s", s);
  int i = 0;
  stack<int> st;
  while (s[i] != '\0') {
    if (s[i] == '(' || s[i] == '{' || s[i] == '[' || s[i] == '<')
      st.push(i);
    else if (st.size() > 0)
      st.pop();
    else {
      printf("Impossible");
      return 0;
    }
    i++;
  }
  if (st.size() != 0) {
    printf("Impossible");
    return 0;
  }
  int ans = 0;
  i = 0;
  stack<char> st1;
  while (s[i] != '\0') {
    if (s[i] == '(' || s[i] == '{' || s[i] == '[' || s[i] == '<')
      st1.push(s[i]);
    else {
      if (s[i] == ')' && st1.top() != '(' || s[i] == '}' && st1.top() != '{' ||
          s[i] == ']' && st1.top() != '[' || s[i] == '>' && st1.top() != '<')
        ans += 1;
      st1.pop();
    }
    i++;
  }
  printf("%d", ans);
  return 0;
}
