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>
constexpr long long inf = (long long)1e9 + 7ll;
char buff[(int)1e6 + 17];
inline void config_io(const bool& __fastio) {
  if (__fastio) {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
  }
  if (!std::string("").empty()) {
    assert(
        freopen(""
                ".in",
                "r", stdin));
    assert(
        freopen(""
                ".out",
                "w", stdout));
  }
}
using namespace std;
const int maxn = (int)2e5 + 17;
int n, ans = 0;
bool open(char c) { return c == '(' || c == '[' || c == '{' || c == '<'; }
bool good(char a, char b) {
  return (a == '(' && b == ')') || (a == '[' && b == ']') ||
         (a == '{' && b == '}') || (a == '<' && b == '>');
}
int main() {
  config_io(false);
  scanf("%s", buff);
  string s = string(buff);
  int bal = 0;
  stack<char> symb;
  for (auto c : s) {
    if (open(c)) {
      ++bal;
      symb.push(c);
    } else {
      if (symb.empty()) return printf("Impossible"), 0;
      if (!good(symb.top(), c)) ++ans;
      symb.pop();
      --bal;
    }
  }
  if (bal != 0)
    printf("Impossible");
  else
    printf("%d\n", ans);
  return 0;
}