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>
using namespace std;
const long long INF = 4e18 + 5;
const int inf = 1e9 + 5;
const int maxn = 1e3 + 5;
bool isClose(char c) {
  if (c == ')' || c == ']' || c == '>' || c == '}') return true;
  return false;
}
bool isMatch(char a, char b) {
  if (a == '(' && b == ')')
    return true;
  else if (a == '[' && b == ']')
    return true;
  else if (a == '{' && b == '}')
    return true;
  else if (a == '<' && b == '>')
    return true;
  return false;
}
string s;
stack<char> stk;
int MAIN() {
  cin >> s;
  int tot = 0;
  for (int i = 0; s[i]; ++i) {
    if (isClose(s[i])) {
      if (stk.empty()) {
        puts("Impossible");
        return 0;
      } else {
        if (!isMatch(stk.top(), s[i])) {
          ++tot;
        }
        stk.pop();
      }
    } else {
      stk.push(s[i]);
    }
  }
  if (stk.empty())
    printf("%d\n", tot);
  else
    puts("Impossible");
  return 0;
}
int main() {
  int ret = MAIN();
  return ret;
}
