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;
int level[1000001], counter = -1, ans = 0;
int code(char x) {
  if (x == '<') {
    return 1;
  } else if (x == '{') {
    return 2;
  } else if (x == '[') {
    return 3;
  } else if (x == '(') {
    return 4;
  } else if (x == '>') {
    return 5;
  } else if (x == '}') {
    return 6;
  } else if (x == ']') {
    return 7;
  } else if (x == ')') {
    return 8;
  }
  return 0;
}
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < s.length(); ++i) {
    int t = code(s[i]);
    if (t <= 4) {
      counter++;
      level[counter] = t;
    } else {
      if (counter == -1) {
        cout << "Impossible";
        return 0;
      } else if (t - level[counter] != 4) {
        ans++;
      }
      counter--;
    }
  }
  if (counter != -1) {
    cout << "Impossible";
    return 0;
  }
  cout << ans;
}
