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>
int main() {
  char s[1000005];
  int i;
  int j;
  int count = 0;
  int prev[1000005];
  int cur = 0;
  int found;
  scanf("%s", s);
  for (i = 0; s[i] != '\0'; i++) {
    if (s[i] == '<' || s[i] == '(' || s[i] == '{' || s[i] == '[') {
      prev[cur] = i;
      cur++;
    } else if (i > 0) {
      cur--;
      if (s[prev[cur]] == '<' && s[i] == '>')
        cur--;
      else if (s[prev[cur]] == '(' && s[i] == ')')
        cur--;
      else if (s[prev[cur]] == '{' && s[i] == '}')
        cur--;
      else if (s[prev[cur]] == '[' && s[i] == ']')
        cur--;
      else {
        cur--;
        count++;
      }
      cur++;
      s[prev[cur]] = 0;
      s[i] = 0;
      if (cur < 0) cur = 0;
    } else {
      printf("Impossible");
      return 0;
    }
  }
  i--;
  while (i >= 0) {
    if (s[i] != 0) {
      printf("Impossible");
      return 0;
    }
    i--;
  }
  printf("%d", count);
  return 0;
}
