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;
stack<char> stk;
char ch;
int fn(char ch) {
  if (ch == '<' || ch == '(' || ch == '[' || ch == '{')
    return 1;
  else
    return 0;
}
int main() {
  int cnt = 0;
  while (scanf("%c", &ch) == 1 && ch != '\n') {
    if (stk.empty()) {
      stk.push(ch);
    } else if (fn(ch) == 1)
      stk.push(ch);
    else {
      if (stk.top() == '<' && ch == '>')
        stk.pop();
      else if (stk.top() == '(' && ch == ')')
        stk.pop();
      else if (stk.top() == '{' && ch == '}')
        stk.pop();
      else if (stk.top() == '[' && ch == ']')
        stk.pop();
      else if (fn(ch) != fn(stk.top())) {
        cnt++;
        stk.pop();
      } else {
        printf("Impossible");
        return 0;
      }
    }
  }
  if (stk.empty()) {
    printf("%d", cnt);
  } else {
    printf("Impossible");
  }
  return 0;
}
