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>
#pragma warning(disable : 4996)
using namespace std;
const long long inf = (int)1e9 - 1;
const int N = ((int)1e5 + 10);
const int K = 70000;
const int mod = 1000000007;
const double eps = 1e-9;
inline void in() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
template <class T>
T modexp(T a, T b, T c) {
  T ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return ans;
}
template <class T>
T C(T n, T r) {
  r = min(r, n - r);
  T ans = 1;
  for (T i = 1; i <= r; i++) {
    ans *= (n - r + i);
    ans /= i;
  }
  return ans;
}
stack<char> st;
int cnt[200];
char f(char x) {
  if (x == '>') return '<';
  if (x == ')') return '(';
  if (x == '}') return '{';
  if (x == ']') return '[';
}
int main() {
  in();
  string s;
  cin >> s;
  int n = ((int)(s).size()), ans = 0;
  for (int i = 0; i <= n - 1; i++) {
    if (st.empty() || s[i] == '<' || s[i] == '{' || s[i] == '[' ||
        s[i] == '(') {
      st.push(s[i]);
      cnt[s[i]]++;
    } else {
      if (f(s[i]) == st.top() && cnt[st.top()] > 0) {
        cnt[st.top()]--;
        st.pop();
      } else {
        char ch = st.top();
        if (ch == '(' || ch == '{' || ch == '[' || ch == '<') {
          cnt[ch]--;
          ans++;
          st.pop();
        } else {
          puts("Impossible");
          return 0;
        }
      }
    }
  }
  st.empty() ? cout << ans : cout << "Impossible";
}
