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>
using namespace std;
string s;
vector<pair<short, int>> v;
int z;
inline short cti(char c) {
  switch (c) {
    case '(':
      return 1;
    case ')':
      return -1;
    case '[':
      return 2;
    case ']':
      return -2;
    case '{':
      return 3;
    case '}':
      return -3;
    case '<':
      return 4;
    case '>':
      return -4;
    default:
      return 0;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  v.reserve(s.size());
  for (size_t i = 0; i < s.size(); i++) {
    short x = cti(s[i]);
    if (x < 0) {
      if (v.empty() || ((v.back().second + i) % 2 == 0)) {
        cout << "Impossible";
        exit(0);
      }
      z += v.back().first != -x;
      v.pop_back();
    } else {
      v.push_back({x, i});
    }
  }
  if (!v.empty()) {
    cout << "Impossible";
    exit(0);
  }
  cout << z;
}