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>
int code(char x) {
  if (x == '{') return 0;
  if (x == '}') return 1;
  if (x == '[') return 2;
  if (x == ']') return 3;
  if (x == '(') return 4;
  if (x == ')') return 5;
  if (x == '<') return 6;
  if (x == '>') return 7;
}
char s[1000005];
int stack[1000005], top = 0;
int n, i, j, k, flag = 1, ans = 0;
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (i = 1; i <= n; i++) {
    if (code(s[i]) & 1) {
      if (top == 0) {
        flag = 0;
        break;
      }
      if (stack[top] != code(s[i]) - 1) ans++;
      top--;
    } else {
      top++;
      stack[top] = code(s[i]);
    }
  }
  if (top) flag = 0;
  if (flag)
    printf("%d", ans);
  else
    printf("Impossible");
  return 0;
}
