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 arr[1000005];
int main() {
  stack<char> mystack;
  int count = 0;
  scanf("%s", arr);
  for (int i = 0; arr[i]; ++i) {
    if (arr[i] == '<' || arr[i] == '(' || arr[i] == '{' || arr[i] == '[')
      mystack.push(arr[i]);
    else {
      if (mystack.empty()) {
        printf("Impossible\n");
        return 0;
      }
      char ch = mystack.top();
      mystack.pop();
      if (arr[i] == '>') {
        if (ch != '<') count++;
      } else if (arr[i] == '}') {
        if (ch != '{') count++;
      } else if (arr[i] == ']') {
        if (ch != '[') count++;
      } else if (arr[i] == ')') {
        if (ch != '(') count++;
      }
    }
  }
  if (!mystack.empty())
    printf("Impossible\n");
  else
    printf("%d\n", count);
  return 0;
}
