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;
int type(char x) {
  if (x == '<' || x == '{' || x == '[' || x == '(') return 0;
  return 1;
}
int fun(char x, char y) {
  if (x == '(' && y == ')') return 1;
  if (x == '<' && y == '>') return 1;
  if (x == '[' && y == ']') return 1;
  if (x == '{' && y == '}') return 1;
  return 0;
}
char t[1000010];
int main() {
  string a;
  while (cin >> a) {
    int j = 0;
    int res = 0;
    bool ok = true;
    for (int i = 0; i < a.length(); i++) {
      if (j > 0) {
        char tmp = t[j];
        if (fun(tmp, a[i])) {
          j--;
          continue;
        }
        if (type(tmp) == 1 && type(a[i]) == 0) {
          ok = false;
          break;
        }
        if (type(a[i]) != type(tmp)) {
          res++;
          j--;
          continue;
        }
        t[++j] = a[i];
      } else
        t[++j] = a[i];
    }
    if (j == 0)
      cout << res << endl;
    else
      cout << "Impossible" << endl;
  }
  return 0;
}
