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 = 1000000 + 10;
char a[maxn];
int top, s[maxn];
int id(char c) {
  if (c == '(') return 0;
  if (c == ')') return 1;
  if (c == '[') return 2;
  if (c == ']') return 3;
  if (c == '{') return 4;
  if (c == '}') return 5;
  if (c == '<') return 6;
  if (c == '>') return 7;
}
int main() {
  scanf("%s", a);
  top = 0;
  int ans = 0;
  for (int i = 0; a[i]; i++) {
    int t = id(a[i]);
    if (!(t & 1)) {
      s[top++] = i;
    } else {
      if (top == 0) {
        printf("Impossible\n");
        return 0;
      }
      top--;
      if ((t ^ id(a[s[top]])) != 1) ans++;
    }
  }
  if (top)
    printf("Impossible\n");
  else
    printf("%d\n", ans);
  return 0;
}
