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;
char str[2000005];
int getID(char c) {
  if (c == '<') return 1;
  if (c == '{') return 2;
  if (c == '[') return 3;
  if (c == '(') return 4;
  if (c == '>') return -1;
  if (c == '}') return -2;
  if (c == ']') return -3;
  if (c == ')') return -4;
  return 0;
}
int st[2000005];
int stop = 0;
int main() {
  scanf("%s", str);
  bool flag = true;
  int change = 0;
  for (int i = 0; str[i] && flag; i++) {
    int id = getID(str[i]);
    if (id > 0) {
      st[stop++] = id;
    } else {
      if (stop <= 0)
        flag = false;
      else {
        int ele = st[--stop];
        if (ele != -id) change++;
      }
    }
  }
  flag = (flag && stop == 0);
  if (!flag)
    puts("Impossible");
  else
    printf("%d\n", change);
  return 0;
}
