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 MAXN = 1e6 + 9;
map<char, int> indices;
int charcount[8];
bool ok(string& s) {
  int count = 0;
  for (int i = 0; i < s.size(); ++i) {
    if (indices[s[i]] > 3) {
      --count;
    } else {
      ++count;
    }
    if (count < 0) {
      return false;
    }
  }
  return count == 0;
}
int main() {
  memset(charcount, 0, 8 * sizeof(int));
  indices['['] = 0;
  indices[']'] = 4;
  indices['<'] = 1;
  indices['>'] = 5;
  indices['{'] = 2;
  indices['}'] = 6;
  indices['('] = 3;
  indices[')'] = 7;
  string s;
  cin >> s;
  if (not ok(s)) {
    cout << "Impossible" << endl;
  } else {
    int ans = 0;
    stack<int> open;
    open.push(indices[s[0]]);
    for (int i = 1; i < s.size(); ++i) {
      int ind = indices[s[i]];
      if (ind < 4)
        open.push(ind);
      else {
        int lastopen = open.top();
        open.pop();
        if (ind != lastopen + 4) ++ans;
      }
    }
    cout << ans << endl;
  }
}
