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;
bool opening(char ch) {
  return ch == '<' || ch == '(' || ch == '[' || ch == '{';
}
char ops(char ch) {
  if (ch == '<') return '>';
  if (ch == '{') return '}';
  if (ch == '[') return ']';
  return ')';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int cnt = 0, good = 1, ans = 0;
  for (char ch : s) {
    if (opening(ch))
      cnt++;
    else
      cnt--;
    if (cnt < 0) good = 0;
  }
  good = (cnt == 0);
  stack<char> st;
  for (char i : s) {
    if (opening(i))
      st.push(i);
    else {
      if (st.empty())
        good = 0;
      else {
        char xx = st.top();
        st.pop();
        if (i != ops(xx)) ans++;
      }
    }
  }
  if (good)
    cout << ans << endl;
  else
    cout << "Impossible" << endl;
}
