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;
template <class T, class U>
void chmin(T &t, U f) {
  if (t > f) t = f;
}
template <class T, class U>
void chmax(T &t, U f) {
  if (t < f) t = f;
}
string s;
char in[] = "[<{(";
char out[] = "]>})";
int main() {
  cin >> s;
  stack<int> stk;
  int ans = 0;
  for (int i = 0; i < (s.size()); i++) {
    int x = find(in, in + 4, s[i]) - in;
    int y = find(out, out + 4, s[i]) - out;
    if (x != 4)
      stk.push(x);
    else {
      if (stk.empty()) {
        cout << "Impossible" << endl;
        return 0;
      }
      if (stk.top() != y) ans++;
      stk.pop();
    }
  }
  if (stk.empty())
    cout << ans << endl;
  else
    cout << "Impossible" << endl;
  return 0;
}
