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 int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
long long add(long long a, long long b, long long mod = MOD) {
  return (a + b) % mod;
}
long long sub(long long a, long long b, long long mod = MOD) {
  return ((a - b) % mod + mod) % mod;
}
long long mul(long long a, long long b, long long mod = MOD) {
  return (a * b) % mod;
}
long long divide(long long a, long long b, long long mod = MOD) {
  return (a * power(b, mod - 2, mod)) % mod;
}
long long sqrt_floor(long long x) {
  long long y = sqrt(x);
  while (y * y > x) --y;
  while ((y + 1) * (y + 1) <= x) ++y;
  return y;
}
int main(int argc, char const *argv[]) {
  string str;
  int cunt = 0;
  cin >> str;
  stack<char> s;
  for (int i = 0; i < str.length(); ++i) {
    if (str[i] == '(' || str[i] == '[' || str[i] == '<' || str[i] == '{') {
      s.push(str[i]);
    } else {
      if (s.empty()) {
        cout << "Impossible\n";
        return 0;
      } else {
        char c = s.top();
        s.pop();
        if (c == '(' && str[i] != ')')
          cunt++;
        else if (c == '[' && str[i] != ']')
          cunt++;
        else if (c == '<' && str[i] != '>')
          cunt++;
        else if (c == '{' && str[i] != '}')
          cunt++;
      }
    }
  }
  if (!s.empty()) {
    cout << "Impossible\n";
    return 0;
  }
  cout << cunt;
  return 0;
}