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;
const long long mod = 1e9 + 7;
const long long MOD = 1e9 + 7;
inline long long add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline long long mul(long long x, long long y) { return (x * 1ll * y) % mod; }
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1e5 + 5;
inline void io() {}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  long long a = 0, ans = 0;
  long long n = s.size();
  for (long long i = 0; i < n; i++) {
    if (s[i] == '{' || s[i] == '<' || s[i] == '(' || s[i] == '[')
      a++;
    else {
      a--;
      if (a < 0) return cout << "Impossible", 0;
    }
  }
  if (a != 0) return cout << "Impossible", 0;
  stack<char> st;
  for (long long i = 0; i < n; i++) {
    if (s[i] == '{' || s[i] == '<' || s[i] == '(' || s[i] == '[')
      st.push(s[i]);
    else {
      if (s[i] == ')') {
        if (st.top() != '(') ans++;
        st.pop();
      }
      if (s[i] == '}') {
        if (st.top() != '{') ans++;
        st.pop();
      }
      if (s[i] == ']') {
        if (st.top() != '[') ans++;
        st.pop();
      }
      if (s[i] == '>') {
        if (st.top() != '<') ans++;
        st.pop();
      }
    }
  }
  cout << ans;
}