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>
struct stack {
  char a[1000010];
  int top;
} st;
void push(char val) { st.a[st.top++] = val; }
int pop() { return st.a[--st.top]; }
int empty() { return st.top == 0; }
char s[1000010];
int main() {
  char t;
  int i, cnt = 0;
  gets(s);
  st.top = 0;
  for (i = 0; s[i]; i++) {
    if (s[i] == '[' || s[i] == '<' || s[i] == '(' || s[i] == '{')
      push(s[i]);
    else {
      if (empty()) {
        printf("Impossible");
        return 0;
      }
      t = pop();
      if (t == '(' && s[i] != ')' || t == '<' && s[i] != '>' ||
          t == '[' && s[i] != ']' || t == '{' && s[i] != '}')
        cnt++;
    }
  }
  if (!empty()) {
    printf("Impossible");
    return 0;
  }
  printf("%d", cnt);
  return 0;
}
