Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const long double eps = 1e-8;
const long double pi = acos(-1);
using namespace std;
char temp[1005];
int ar[] = {0, 0, -1, 1};
int ac[] = {-1, 1, 0, 0};
char opened[] = {'[', '{', '<', '('};
char closed[] = {']', '}', '>', ')'};
bool open(char c) { return c == '[' || c == '{' || c == '(' || c == '<'; }
bool close(char c) { return !open(c); }
void finish() {
  cout << "Impossible\n";
  exit(0);
}
char a[1005][1005];
int d[1005][1005];
int n, m;
bool ok(int r, int c) { return r >= 0 && c >= 0 && r < n && c < m; }
struct node {
  int r, c, w;
  node(int r, int c, int w) : r(r), c(c), w(w) {}
  bool operator<(const node &n) const { return this->w > n.w; }
};
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  stack<char> st;
  map<char, char> mp;
  mp['['] = ']', mp[']'] = '[';
  mp['{'] = '}', mp['}'] = '{';
  mp['<'] = '>', mp['>'] = '<';
  mp['('] = ')', mp[')'] = '(';
  int res = 0;
  for (int i = 0; i < s.size(); i++) {
    if (open(s[i])) st.push(s[i]);
    if (close(s[i])) {
      if (st.empty()) finish();
      char c = st.top();
      st.pop();
      if (mp[c] - s[i]) res++;
      s[i] = mp[c];
    }
  }
  if (!st.empty()) finish();
  cout << res << "\n";
  return 0;
}