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;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
char s[1000010];
int deg, res, val[100000];
char fil[1000010];
int main() {
  scanf(" %s", s);
  val['('] = 0;
  val[')'] = 1;
  val['['] = 2;
  val[']'] = 3;
  val['{'] = 4;
  val['}'] = 5;
  val['<'] = 6;
  val['>'] = 7;
  int tam = strlen(s);
  res = deg = 0;
  for (int a = tam - 1; a >= 0; a--) {
    if (s[a] == ']' || s[a] == ')' || s[a] == '}' || s[a] == '>') {
      fil[deg++] = s[a];
    } else {
      if (deg == 0) {
        printf("Impossible\n");
        return 0;
      }
      if (val[fil[deg - 1]] != val[s[a]] + 1) res++;
      deg--;
    }
  }
  if (deg != 0) {
    printf("Impossible\n");
    return 0;
  }
  printf("%d\n", res);
}
