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 double pi = acos(-1.0);
const long long INF = 2e18;
void print(int a[], int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
void printll(long long a[], long long n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
vector<long long> findPrimeFactors(long long n) {
  vector<long long> primeFactors;
  while (n % 2 == 0) primeFactors.push_back(2), n = n / 2;
  for (long long i = 3; i * i <= n; i = i + 2) {
    while (n % i == 0) {
      primeFactors.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) primeFactors.push_back(n);
  return primeFactors;
}
vector<long long> allDivisiors(long long n) {
  vector<long long> divisors;
  long long limit = sqrt(n + 1);
  for (long long i = 1; i <= limit; i++) {
    if (n % i == 0) {
      if (n / i == i)
        divisors.push_back(i);
      else
        divisors.push_back(i), divisors.push_back(n / i);
    }
  }
  return divisors;
}
bool isPrime(long long n) {
  if (n == 2 or n == 3) return true;
  if (n % 2 == 0) return false;
  for (long long i = 3; i * i <= n; i += 2)
    if (n % i == 0) return false;
  return true;
}
long long nsum(long long n) { return (n * (n + 1)) / 2; }
long long power(long long n, long long p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * n;
    n = n * n;
    p = p / 2;
  }
  return ans;
}
long long mpower(long long n, long long p, long long mod) {
  if (p == -1) p = mod - 2;
  if (p == 0) return 1;
  if (p == 1) return n;
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    p = p / 2;
  }
  return ans;
}
bool isPalindrome(string s) {
  int N = s.size();
  for (int i = 0; i < (N / 2); i++) {
    if (s[i] != s[N - 1 - i]) return 0;
  }
  return 1;
}
long long mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
jump:
  while (tc--) {
    string s;
    cin >> s;
    long long f = 0;
    long long ok = 1;
    for (auto t : s) {
      if (t == '<' or t == '{' or t == '(' or t == '[')
        f++;
      else
        f--;
      if (f == -1) {
        ok = 0;
        break;
      }
    }
    if (f != 0) ok = 0;
    if (ok == 0)
      cout << "Impossible";
    else {
      stack<char> ST;
      long long ans = 0;
      for (auto t : s) {
        if (t == '<' or t == '{' or t == '(' or t == '[')
          ST.push(t);
        else {
          char c = ST.top();
          if ((c == '<' and t == '>') or (c == '(' and t == ')') or
              (c == '{' and t == '}') or (c == '[' and t == ']'))
            ST.pop();
          else
            ans++, ST.pop();
        }
      }
      cout << ans;
    }
  }
}