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;
string S;
map<char, int> conv;
int main() {
  conv['<'] = 1;
  conv['{'] = 2;
  conv['('] = 3;
  conv['['] = 4;
  conv['>'] = -1;
  conv['}'] = -2;
  conv[')'] = -3;
  conv[']'] = -4;
  cin >> S;
  stack<char> stk;
  int ans = 0;
  for (char ch : S) {
    if (ch == '<' || ch == '{' || ch == '(' || ch == '[')
      stk.push(ch);
    else if (stk.empty()) {
      printf("Impossible\n");
      return 0;
    } else if (conv[stk.top()] + conv[ch] == 0) {
      stk.pop();
    } else {
      ans++;
      stk.pop();
    }
  }
  if (!stk.empty()) {
    printf("Impossible\n");
  } else {
    printf("%d\n", ans);
  }
  return 0;
}
