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;
const long long MAXVAL = 1000000000000000000LL;
const long long MINVAL = -1000000000000000000LL;
double sqr(double a) { return a * a; }
string s;
stack<char> st;
int res = 0;
char a = '{';
char aa = '}';
char b = '[';
char bb = ']';
char c = '<';
char cc = '>';
char d = '(';
char dd = ')';
int main() {
  cin >> s;
  for (int i = 0; i < (int)s.length(); ++i) {
    if (s[i] == '<' || s[i] == '(' || s[i] == '{' || s[i] == '[') {
      st.push(s[i]);
      continue;
    }
    if (st.empty()) {
      cout << "Impossible" << endl;
      return 0;
    }
    if ((s[i] == aa && st.top() == a) || (s[i] == bb && st.top() == b) ||
        (s[i] == cc && st.top() == c) || (s[i] == dd && st.top() == d)) {
      st.pop();
      continue;
    } else {
      st.pop();
      res++;
    }
  }
  if (!st.empty()) {
    cout << "Impossible" << endl;
    return 0;
  }
  cout << res << endl;
  return 0;
}
