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>
const int MOD = 1000000007;
using namespace std;
void solve() {
  string s;
  cin >> s;
  stack<char> st;
  int cl = 0, op = 0;
  for (auto x : s) {
    if (x == '<' || x == '(' || x == '[' || x == '{')
      op++;
    else
      cl++;
  }
  if (op != cl) {
    cout << "Impossible";
    return;
  }
  int ans = 0;
  for (auto x : s) {
    if (st.empty()) {
      if (x == '<' || x == '(' || x == '[' || x == '{')
        st.push(x);
      else {
        cout << "Impossible";
        return;
      }
    } else if (x == '<' || x == '(' || x == '[' || x == '{') {
      st.push(x);
    } else if (x == '>') {
      if (st.top() == '<') {
        st.pop();
      } else if (st.top() == '(' || st.top() == '[' || st.top() == '{') {
        ans++;
        st.pop();
      }
    } else if (x == ')') {
      if (st.top() == '(') {
        st.pop();
      } else if (st.top() == '<' || st.top() == '[' || st.top() == '{') {
        ans++;
        st.pop();
      }
    } else if (x == '}') {
      if (st.top() == '{') {
        st.pop();
      } else if (st.top() == '<' || st.top() == '[' || st.top() == '(') {
        ans++;
        st.pop();
      }
    } else if (x == ']') {
      if (st.top() == '[') {
        st.pop();
      } else if (st.top() == '(' || st.top() == '<' || st.top() == '{') {
        ans++;
        st.pop();
      }
    }
  }
  cout << ans;
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
    cout << endl;
  }
}
