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;
using ll = long long int;
using vll = vector<ll>;
using vii = vector<int>;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
using ull = unsigned long long int;
ll To_int(string str) {
  stringstream ss(str);
  ll x = 0;
  ss >> x;
  return x;
}
string To_string(ll x) {
  stringstream ss;
  ss << x;
  string str = ss.str();
  return str;
}
template <typename T>
T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
void inout() {}
bool isopening(char ch) {
  if (ch == '(') return true;
  if (ch == '{') return true;
  if (ch == '[') return true;
  if (ch == '<') return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  stack<char> st;
  bool ok = true;
  char TOP;
  int cnt = 0;
  for (int i = 0; i < s.size(); i++) {
    if (isopening(s[i]))
      st.push(s[i]);
    else if (st.empty()) {
      cout << "Impossible";
      return 0;
    } else {
      TOP = st.top();
      st.pop();
      if (TOP == '(' and s[i] == ')')
        continue;
      else if (TOP == '{' and s[i] == '}')
        continue;
      else if (TOP == '[' and s[i] == ']')
        continue;
      else if (TOP == '<' and s[i] == '>')
        continue;
      else
        cnt++;
    }
  }
  if (!st.empty())
    cout << "Impossible";
  else
    cout << cnt;
  return 0;
}