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;
int main() {
  int r = 0, x, y;
  r = 0;
  stack<int> s;
  char c;
  while (cin.get(c)) {
    switch (c) {
      case '[':
        x = -1;
        break;
      case ']':
        x = 1;
        break;
      case '{':
        x = -2;
        break;
      case '}':
        x = 2;
        break;
      case '(':
        x = -3;
        break;
      case ')':
        x = 3;
        break;
      case '<':
        x = -4;
        break;
      case '>':
        x = 4;
        break;
      default:
        continue;
    }
    if (s.size() > 0) {
      y = s.top();
      if (y > 0 || x < 0) {
        s.push(x);
        continue;
      }
      s.pop();
      if (x + y != 0) r++;
    } else
      s.push(x);
  }
  if (s.size() > 0)
    cout << "Impossible";
  else
    cout << r << endl << flush;
  return 0;
}
