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;
const int Maxn = 1e6 + 5;
int main() {
  char c, a[Maxn];
  int cnt = 0, ans = 0;
  do {
    c = getchar();
    if (c == '<' || c == '(' || c == '{' || c == '[') {
      a[cnt] = c;
      cnt++;
    }
    if (c == '>' || c == ')' || c == '}' || c == ']') {
      cnt--;
      ans++;
      if (a[cnt] == '<' && c == '>') ans--;
      if (a[cnt] == '(' && c == ')') ans--;
      if (a[cnt] == '{' && c == '}') ans--;
      if (a[cnt] == '[' && c == ']') ans--;
    }
    if (cnt < 0) c = '\n';
  } while (c != '\n');
  if (cnt != 0)
    printf("Impossible\n");
  else
    printf("%d\n", ans);
  return 0;
}
