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;
char s[1000555], Stack[1000555];
int top, Ans;
int main() {
  fgets(s, 1000005, stdin);
  top = 0;
  Ans = 0;
  for (int i = 0; s[i] != '\n'; i++) {
    if (s[i] == '{')
      Stack[++top] = '{';
    else if (s[i] == '[')
      Stack[++top] = '[';
    else if (s[i] == '<')
      Stack[++top] = '<';
    else if (s[i] == '(')
      Stack[++top] = '(';
    else {
      if (top == 0) {
        puts("Impossible");
        return 0;
      }
      if (s[i] == '}') {
        if (Stack[top] != '{') Ans++;
      } else if (s[i] == ']') {
        if (Stack[top] != '[') Ans++;
      } else if (s[i] == ')') {
        if (Stack[top] != '(') Ans++;
      } else if (s[i] == '>') {
        if (Stack[top] != '<') Ans++;
      }
      top--;
    }
  }
  if (top > 0) {
    puts("Impossible");
    return 0;
  }
  printf("%d\n", Ans);
}
