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;
long long i, res = 0, a[1000001];
stack<long long> st;
string s;
int main() {
  cin >> s;
  for (i = 0; i < s.size(); i++) {
    if (s[i] == '<') a[i] = 1;
    if (s[i] == '>') a[i] = 11;
    if (s[i] == '{') a[i] = 2;
    if (s[i] == '}') a[i] = 22;
    if (s[i] == '(') a[i] = 3;
    if (s[i] == ')') a[i] = 33;
    if (s[i] == '[') a[i] = 4;
    if (s[i] == ']') a[i] = 44;
  }
  for (i = 0; i < s.size(); i++) {
    if (a[i] == 1 || a[i] == 2 || a[i] == 3 || a[i] == 4)
      st.push(a[i]);
    else {
      if (st.empty()) {
        cout << "Impossible";
        return 0;
      }
      if ((st.top() * 10) + st.top() != a[i]) res++;
      st.pop();
    }
  }
  if (!st.empty()) {
    cout << "Impossible";
    return 0;
  }
  cout << res;
  return 0;
}
