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
/**
 *  CodeForces/312 C. Replace To Make Regular Bracket Sequence
 *  
 *  @author Brooks Mershon
 *
 */

// return Impossible or the least amount of exchanges to fix bracket sequence
function replace(x) {
  var N = x.length,
      count = 0,
      invalid = "Impossible";

  var open  = ['<', '{', '[', '('],
      close = ['>', '}', ']', ')'];

  // scan to check whether the sequence is balanced
  var d = 0;
  for(var i = 0; i < N; i++) {
    d += (open.indexOf(x.charAt(i)) > -1) ? 1 : -1;
    if (d < 0) return invalid;
  }

  if (d !== 0) return invalid;

  var q = [];
  // greedy search for matching parenthesis
  var k = -1;
  while(k < x.length - 1) {
    while(open.indexOf(x.charAt(++k)) > -1) { // iterate to next closing bracket
      q.push(x.charAt(k));
    }
    var opener = q.pop();
    if (open.indexOf(opener) !== close.indexOf(x.charAt(k))) count++;
  }

  return count;
}

// Codeforces I/O
print(replace(readline()));

//module.exports = replace;
