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>
using namespace std;
const double pi = 4.0 * atan(1.0);
const int N = 10000000;
char *str;
stack<int> stc;
int convert(char ch) {
  switch (ch) {
    case '<':
      return 1;
    case '(':
      return 2;
    case '[':
      return 3;
    case '{':
      return 4;
    case '>':
      return 5;
    case ')':
      return 6;
    case ']':
      return 7;
    case '}':
      return 8;
  }
}
void init() { str = (char *)malloc(N * sizeof(char)); }
int main() {
  int T, cnt, flag, tmp, changes;
  flag = 1;
  changes = 0;
  init();
  scanf("%s", str);
  cnt = 0;
  while (str[cnt] != '\0') {
    tmp = convert(str[cnt]);
    if (tmp <= 4)
      stc.push(tmp);
    else {
      if (stc.size() == 0) {
        flag = 0;
        break;
      }
      if (stc.top() >= 5) {
        flag = 0;
        break;
      } else if (tmp - 4 == stc.top())
        stc.pop();
      else if (tmp - 4 != stc.top()) {
        stc.pop();
        changes++;
      }
    }
    cnt++;
  }
  if (!flag || stc.size() != 0)
    printf("Impossible\n");
  else
    printf("%d\n", changes);
}