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> rbs;
int main() {
  string s;
  cin >> s;
  int ans = 0;
  for (int(i) = 0; (i) < ((int)(s.size())); (i)++) {
    if (s[i] == '<' or s[i] == '{' or s[i] == '(' or s[i] == '[')
      rbs.push(s[i]);
    else {
      if (rbs.empty()) {
        ans = -1;
        break;
      }
      char c = rbs.top();
      rbs.pop();
      if ((c == '{' and s[i] != '}') or (c == '<' and s[i] != '>') or
          (c == '(' and s[i] != ')') or (c == '[' and s[i] != ']'))
        ans += 1;
    }
  }
  if (ans != -1 and rbs.empty())
    printf("%d\n", ans);
  else
    printf("Impossible");
  return 0;
}
