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;
string s;
char v[1000005];
int o, a, n;
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) {
    char c = s[i];
    if (c == '(' || c == '[' || c == '<' || c == '{')
      a++;
    else
      a--;
    if (a < 0) a = -123456;
  }
  if (a) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    char c = s[i];
    if (c == '<' || c == '{' || c == '(' || c == '[') {
      v[++o] = c;
      continue;
    }
    if (c == '}')
      c = '{';
    else if (c == '>')
      c = '<';
    else if (c == ')')
      c = '(';
    else
      c = '[';
    if (v[o--] != c) a++;
  }
  printf("%d\n", a);
  return 0;
}
