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;
bool result(char x) { return (x == '(' || x == '<' || x == '{' || x == '['); }
char result1(char x) {
  if (x == '>') return '<';
  if (x == ']') return '[';
  if (x == '}') return '{';
  if (x == ')') return '(';
}
int main() {
  stack<char> stk;
  string x;
  int res = 0;
  cin >> x;
  for (int i = 0; i < x.length(); i++) {
    if (result(x[i])) {
      stk.push(x[i]);
    } else {
      if (i == 0) {
        stk.push(x[i]);
        break;
      }
      if (stk.empty()) {
        cout << "Impossible" << endl;
        return 0;
      }
      char a = result1(x[i]);
      char b = stk.top();
      if (b != a) {
        res++;
      }
      stk.pop();
    }
  }
  if (stk.empty()) {
    cout << res;
  } else {
    cout << "Impossible" << endl;
    return 0;
  }
}
