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 int MAX_N = 1000005;
char s[MAX_N];
const string Impossible = "Impossible";
stack<char> st;
inline void check() {
  if (st.empty()) {
    printf("%s\n", Impossible.c_str());
    exit(0);
  }
}
int main() {
  scanf("%s", s);
  int n = strlen(s);
  int res = 0;
  for (int i = 0; i < (int)n; ++i) {
    if (s[i] == ')') {
      check();
      if (st.top() != '(') ++res;
      st.pop();
    } else if (s[i] == ']') {
      check();
      if (st.top() != '[') ++res;
      st.pop();
    } else if (s[i] == '>') {
      check();
      if (st.top() != '<') ++res;
      st.pop();
    } else if (s[i] == '}') {
      check();
      if (st.top() != '{') ++res;
      st.pop();
    } else
      st.push(s[i]);
  }
  if (!st.empty())
    printf("%s\n", Impossible.c_str());
  else
    printf("%d\n", res);
  return 0;
}
