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;
using vi = vector<int>;
using pii = pair<int, int>;
using ull = unsigned long long;
using ll = long long;
bool check(char c, char c2) {
  return (c == '(' && c2 == ')') || (c == '<' && c2 == '>') ||
         (c == '[' && c2 == ']') || (c == '{' && c2 == '}');
}
int solve(const string &str) {
  stack<char> s;
  int cnt = 0;
  for (const auto i : str) {
    if (i == '(' || i == '[' || i == '{' || i == '<') {
      s.push(i);
    }
    if (i == ')' || i == ']' || i == '}' || i == '>') {
      if (s.empty()) {
        return -1;
      } else {
        if (!check(s.top(), i)) cnt++;
        s.pop();
      }
    }
  }
  if (s.empty())
    return cnt;
  else
    return -1;
}
int main() {
  string s;
  cin >> s;
  int ans = solve(s);
  if (ans == -1)
    cout << "Impossible\n";
  else
    cout << ans << "\n";
}
