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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
char str[1000000 + 10];
int change(char c) {
  switch (c) {
    case '<':
      return 1;
    case '>':
      return -1;
    case '{':
      return 2;
    case '}':
      return -2;
    case '[':
      return 3;
    case ']':
      return -3;
    case '(':
      return 4;
    default:
      return -4;
  }
}
int main() {
  scanf("%s", str);
  int len = strlen(str);
  stack<int> s;
  int sum = 0;
  int flag = 0;
  for (int i = 0; i < len; i++) {
    int temp = change(str[i]);
    if (!s.empty()) {
      if (temp > 0)
        s.push(temp);
      else {
        if (temp + s.top() != 0) {
          sum++;
          s.pop();
        } else
          s.pop();
      }
    } else {
      if (temp > 0)
        s.push(temp);
      else {
        flag = 1;
      }
    }
  }
  if ((flag) || (!s.empty()))
    printf("Impossible\n");
  else
    printf("%d\n", sum);
  return 0;
}
