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 N = 1e6 + 2;
char s[N], st[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  int r = 0, ans = 0;
  bool ok = true;
  for (int i = 1; i < n + 1; ++i) {
    if (s[i] == '(' || s[i] == '<' || s[i] == '{' || s[i] == '[')
      st[++r] = s[i];
    else {
      if (r < 1) return puts("Impossible"), 0;
      if (st[r] == '(' && s[i] != ')') ans++;
      if (st[r] == '[' && s[i] != ']') ans++;
      if (st[r] == '<' && s[i] != '>') ans++;
      if (st[r] == '{' && s[i] != '}') ans++;
      r--;
    }
  }
  if (!r)
    printf("%d\n", ans);
  else
    puts("Impossible");
  return 0;
}
