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;
string open = "([{<";
string closed = ")]}>";
string st;
int cont = 0;
void RBS(int i, string &s) {
  if (i == st.size()) {
    if (s.size() > 0) cont = -1;
    return;
  }
  if (s.empty() and -1 == open.find(st[i])) {
    cont = -1;
    return;
  }
  unsigned int pos = open.find(st[i]);
  if (pos != -1)
    s += st[i], RBS(i + 1, s);
  else {
    pos = closed.find(st[i]);
    if (*s.rbegin() != open[pos]) ++cont;
    s.erase(s.end() - 1);
    RBS(i + 1, s);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s = "";
  cin >> st;
  RBS(0, s);
  if (cont < 0)
    cout << "Impossible"
         << "\n";
  else
    cout << cont << "\n";
  return 0;
}