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 MAXN = 1e6 + 7;
char s[MAXN];
int A[MAXN];
int n;
string brackets = "<{[( )]}>";
void load() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    A[i] = brackets.find(s[i]) - 4;
  }
}
bool solve() {
  stack<int> S;
  int sol = 0;
  for (int i = 0; i < n; i++) {
    if (A[i] < 0)
      S.push(A[i]);
    else {
      if (S.empty()) return false;
      if (S.top() != -A[i]) sol++;
      S.pop();
    }
  }
  if (!S.empty()) return false;
  printf("%d\n", sol);
  return true;
}
int main() {
  load();
  if (!solve()) printf("Impossible\n");
  return 0;
}
