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 double PI = acos(-1);
pair<int, int> brac[4];
map<char, int> toInt = {{'<', 0}, {'>', 0}, {'(', 1}, {')', 1},
                        {'{', 2}, {'}', 2}, {'[', 3}, {']', 3}};
int main() {
  char s[1000005];
  scanf("%s", s);
  int len = strlen(s);
  int diff = 0;
  stack<int> br;
  for (auto i = 0; i < len; ++i) {
    if (s[i] == '<' || s[i] == '(' || s[i] == '{' || s[i] == '[') {
      br.push(toInt[s[i]]);
    } else {
      if (br.empty()) {
        printf("Impossible\n");
        return 0;
      }
      if (br.top() != toInt[s[i]]) diff++;
      br.pop();
    }
  }
  if (!br.empty()) {
    printf("Impossible\n");
    return 0;
  }
  printf("%d\n", diff);
  return 0;
}
