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>
using namespace std;
#pragma GCC optimize("-O3")
const long double PI = 3.141592653589793;
const long long int inf = 1000000000000000000;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
vector<string> splitstring(string str) {
  stringstream ss(str);
  vector<string> s;
  string word;
  while (ss >> word) {
    s.push_back(word);
  }
  return s;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, i, cf = 0, cb = 0, cc = 0, cnt = 0, count = 0;
  string s;
  cin >> s;
  n = s.size();
  stack<char> st;
  for (i = 0; i < n; i++) {
    if (((s[i] == ']') || (s[i] == '>') || (s[i] == '>') || (s[i] == '}')) and
        cc == 0) {
      cc--;
    }
    if ((s[i] == '[') || (s[i] == '<') || (s[i] == '(') || (s[i] == '{')) {
      st.push(s[i]);
      cc++;
    } else if (cc > 0 and ((s[i] == ']') || (s[i] == '>') || (s[i] == ')') ||
                           (s[i] == '}'))) {
      char ch = st.top();
      if ((ch == '(' and s[i] == ')') || (ch == '[' and s[i] == ']') ||
          (ch == '{' and s[i] == '}') || (ch == '<' and s[i] == '>')) {
        st.pop();
        cnt++;
        cc--;
      } else {
        cb++;
        count++;
        st.pop();
        cc--;
      }
    }
    if (cc < 0) break;
  }
  if (i == n) {
    if (cc > 0) {
      cout << "Impossible";
    } else {
      cout << count;
    }
  } else
    cout << "Impossible";
  return 0;
}