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 int N = 2e5 + 1;
int main() {
  int admit = 0;
  int oBlock = 0, cBlock = 0;
  stack<char> q1;
  int ans = 0;
  for (char c = getchar(); c != '\n';) {
    if (c == '<' || c == '(' || c == '[' || c == '{') {
      admit++;
      switch (c) {
        case '<': {
          q1.push('>');
          break;
        }
        case '(': {
          q1.push(')');
          break;
        }
        case '[': {
          q1.push(']');
          break;
        }
        case '{': {
          q1.push('}');
          break;
        }
      }
    } else {
      admit--;
      if (admit < 0) {
        cout << "Impossible" << endl;
        return 0;
      }
      if (c != q1.top()) ans++;
      q1.pop();
    }
    c = getchar();
  }
  if (admit != 0) {
    cout << "Impossible" << endl;
    return 0;
  }
  printf("%d", ans);
  return 0;
}
