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;
bool ch(char x, char y) {
  if (x == '{' && y == '}' || x == '[' && y == ']' || x == '<' && y == '>' ||
      x == '(' && y == ')')
    return 1;
  return 0;
}
int main() {
  string s;
  cin >> s;
  vector<char> open = {'(', '{', '[', '<'};
  vector<char> close = {')', '}', ']', '>'};
  int r = 0;
  for (int i = 0; i < s.size(); i++) {
    if ((find(open.begin(), open.end(), s[i]) == open.end())) {
      r--;
    } else {
      r++;
    }
    if (r < 0) {
      cout << "Impossible";
      return 0;
    }
  }
  if (r) {
    cout << "Impossible";
    return 0;
  }
  int ans = 0;
  vector<char> v;
  for (int i = 0; i < s.size(); i++) {
    if (find(open.begin(), open.end(), s[i]) != open.end()) {
      v.push_back(s[i]);
    } else {
      if (!ch(v.back(), s[i])) {
        ans++;
      }
      v.pop_back();
    }
  }
  cout << ans;
  return 0;
}
