Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int mod_pow(long long int a, long long int n, long long int b) {
  long long int res = 1;
  while (n) {
    if (n & 1) res = (res * a) % b;
    a = (a * a) % b;
    n /= 2;
  }
  return res % b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  stack<char> rbs;
  string s;
  cin >> s;
  char a[4] = {'(', '{', '[', '<'};
  long long int sz = s.length();
  long long int left = 0, right = 0;
  long long int ans = 0;
  for (long long int i = 0; i < sz; i++) {
    if (s[i] == '(' || s[i] == '[' || s[i] == '<' || s[i] == '{') {
      rbs.push(s[i]);
      ++left;
    } else {
      ++right;
      if (right > left) {
        cout << "Impossible";
        return 0;
      }
      long long int pos;
      if (s[i] == ')') pos = 0;
      if (s[i] == '}') pos = 1;
      if (s[i] == ']') pos = 2;
      if (s[i] == '>') pos = 3;
      if (rbs.top() != a[pos]) {
        ans++;
      }
      rbs.pop();
    }
  }
  if (left > right) {
    cout << "Impossible";
    return 0;
  }
  cout << ans;
  return 0;
}