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;
int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000000;
    y = y >> 1;
    x = (x * x) % 1000000000;
  }
  return res;
}
void solve() {
  string s;
  cin >> s;
  stack<char> st;
  int f = 0;
  int n = (int)s.length();
  int ans = 0, cnt = 0;
  for (int i = 0; i < n; i++) {
    if ((s[i] == ')' || s[i] == '}' || s[i] == '>' || s[i] == ']') &&
        cnt <= 0) {
      f = 1;
      break;
    } else {
      if (s[i] == '(' || s[i] == '{' || s[i] == '<' || s[i] == '[') {
        st.push(s[i]);
        cnt++;
      } else {
        cnt--;
        char c = st.top();
        st.pop();
        if ((s[i] == ')' && c == '(') || (s[i] == '}' && c == '{') ||
            (s[i] == '>' && c == '<') || (s[i] == ']' && c == '['))
          continue;
        else
          ans++;
      }
    }
  }
  if (!st.empty()) f = 1;
  if (f)
    cout << "Impossible\n";
  else
    cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  while (T--) solve();
}
