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 int inf = 0x3f3f3f3f, mod = 1000000007;
const double pi = 3.1415926535897932, eps = 1e-6;
string s;
int st[1000005], n, t, m[128], ans;
int main() {
  m['('] = 1;
  m['<'] = 2;
  m['['] = 3;
  m['{'] = 4;
  m[')'] = 5;
  m['>'] = 6;
  m[']'] = 7;
  m['}'] = 8;
  cin >> s;
  n = s.size();
  for (int(i) = (0); (i) <= (n - 1); (i)++) {
    if (m[s[i]] < 5)
      st[++t] = m[s[i]];
    else {
      if (t <= 0) {
        cout << "Impossible";
        return 0;
      }
      ans += (abs(st[t] - m[s[i]]) != 4);
      t--;
    }
  }
  if (t > 0) {
    cout << "Impossible";
    return 0;
  }
  cout << ans;
  return 0;
}
