Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int openind[] = {40, 60, 91, 123};
  int closeind[] = {41, 62, 93, 125};
  stack<int> s1;
  string s;
  cin >> s;
  int n = s.length();
  int o = 0;
  long long res1 = 0;
  for (int i = 0; i < n; i++) {
    bool b = false;
    for (int j = 0; j < 4; j++) {
      if (s[i] == openind[j]) {
        o++;
        s1.push(j);
        b = true;
        break;
      }
    }
    if (!b) {
      if (!s1.empty()) {
        int t = s1.top();
        s1.pop();
        if (closeind[t] != s[i]) {
          res1++;
        }
      } else {
        printf("Impossible");
        exit(0);
      }
    }
  }
  if (!s1.empty())
    printf("Impossible");
  else
    printf("%I64d", res1);
  return 0;
}