Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
char s[N];
stack<char> lol;
map<char, char> sweet;
int main() {
  scanf("%s", s);
  string ok = s, sh = ")]>}", sh2 = "([<{";
  int os = 0;
  for (int i = 0; i < 4; i++) sweet[sh2[i]] = sh[i];
  for (int i = 0; i < ok.size(); i++) {
    if (lol.size() == 0 && sh.find(ok[i]) != -1)
      return printf("Impossible"), 0;
    else if (sh2.find(ok[i]) != -1)
      lol.push(ok[i]);
    else {
      os += ((sweet[lol.top()] != ok[i]) ? 1 : 0);
      lol.pop();
    }
  }
  if (lol.size() == 0)
    printf("%d", os);
  else
    printf("Impossible");
  return 0;
}