Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
typedef struct link {
  char element;
  struct link* next;
} Link;
typedef struct stack {
  Link* top;
  int size;
} Stack;
Link* create_link(char it, Link* nextval) {
  Link* n = (Link*)malloc(sizeof(Link));
  n->element = it;
  n->next = nextval;
  return n;
}
Stack* create_stack() {
  Stack* s = (Stack*)malloc(sizeof(Stack));
  s->top = NULL;
  s->size = 0;
  return s;
}
void push(Stack* s, char it) {
  s->top = create_link(it, s->top);
  s->size++;
}
void pop(Stack* s) {
  if (s->top == NULL) return;
  s->top = s->top->next;
  s->size--;
}
char topValue(Stack* s) {
  if (s->top == NULL) return -1;
  return s->top->element;
}
int main() {
  char braket;
  int count = 0;
  Stack* atual;
  int open = 0, close = 0;
  char retornado;
  int tick = 0;
  atual = create_stack();
  scanf("%c", &braket);
  if (braket == '}' || braket == ']' || braket == '>' || braket == ')')
    printf("Impossible");
  else {
    while (braket != '\n') {
      if (braket == '{' || braket == '[' || braket == '<' || braket == '(') {
        push(atual, braket);
        open++;
      } else {
        close++;
        retornado = topValue(atual);
        if (retornado != '[' && retornado != '{' && retornado != '(' &&
            retornado != '<')
          close = close + 352;
        if ((retornado == '{' && braket == '}') ||
            (retornado == '[' && braket == ']') ||
            (retornado == '<' && braket == '>') ||
            (retornado == '(' && braket == ')'))
          pop(atual);
        else {
          count++;
          pop(atual);
        }
      }
      scanf("%c", &braket);
    }
    if (open != close)
      printf("Impossible");
    else
      printf("%d", count);
  }
  return 0;
}