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;
char c;
int name;
stack<char> s;
bool check1(char ch) {
  return (ch == '[' || ch == '<' || ch == '{' || ch == '(');
}
bool check2(char ch) {
  return (ch == ']' || ch == '>' || ch == '}' || ch == ')');
}
int main() {
  int ans = 0;
  while (scanf("%c", &c) != EOF) {
    if (!check1(c) && !check2(c)) break;
    if (check1(c))
      s.push(c);
    else {
      if (s.empty()) {
        printf("Impossible");
        return 0;
      }
      char top = s.top();
      s.pop();
      if (top == '(' && c != ')')
        ans++;
      else if (top == '{' && c != '}')
        ans++;
      else if (top == '<' && c != '>')
        ans++;
      else if (top == '[' && c != ']')
        ans++;
    }
  }
  if (!s.empty())
    printf("Impossible");
  else
    printf("%d", ans);
  return 0;
}
