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 INF = 0x3f3f3f3f;
const long long int llINF = 0x3f3f3f3f3f3f3f;
bool match(char x, char y) {
  if ((x == '{' && y == '}') || (y == '{' && x == '}') ||
      (x == '(' && y == ')') || (y == '(' && x == ')'))
    return true;
  ;
  if ((x == '[' && y == ']') || (y == '[' && x == ']') ||
      (x == '<' && y == '>') || (y == '<' && x == '>'))
    return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  int cost = 0;
  stack<char> pilha;
  for (auto x : s) {
    if (x == '(' || x == '{' || x == '[' || x == '<') {
      pilha.push(x);
      continue;
    } else {
      if (pilha.empty()) {
        puts("Impossible");
        return 0;
      }
      char aux = pilha.top();
      pilha.pop();
      if (!match(x, aux)) cost++;
    }
  }
  if (!pilha.empty()) {
    puts("Impossible");
    return 0;
  }
  cout << cost << endl;
}
