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 MOD = 1e9 + 7;
const int INF = 1e9 + 9;
const int MX = 1e6 + 5;
char inp[MX];
int main() {
  cin >> inp;
  int l = strlen(inp);
  if (inp[0] == ')' or inp[0] == '}' or inp[0] == ']' or inp[0] == '>') {
    cout << "Impossible" << endl;
    return 0;
  }
  int op = 0;
  stack<char> st;
  st.push(inp[0]);
  for (int i = 1; i < l; i++) {
    if (inp[i] == '(' or inp[i] == '{' or inp[i] == '[' or inp[i] == '<')
      st.push(inp[i]);
    else {
      if (st.empty()) {
        cout << "Impossible" << endl;
        return 0;
      }
      char c = st.top();
      st.pop();
      if (inp[i] == ')' && c != '(') op++;
      if (inp[i] == '}' && c != '{') op++;
      if (inp[i] == ']' && c != '[') op++;
      if (inp[i] == '>' && c != '<') op++;
    }
  }
  if (!st.empty())
    cout << "Impossible" << endl;
  else
    cout << op << endl;
  return 0;
}
