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;
#pragma warning(disable : 4996)
using namespace std;
char a[1000000 + 10];
int main() {
  int i, k;
  while (scanf("%s", a) != EOF) {
    int len = strlen(a);
    if (len & 1) {
      printf("Impossible\n");
      continue;
    }
    stack<char> s;
    bool flag = true;
    k = 0;
    for (i = 0; i < len; i++) {
      if (a[i] == '<' || a[i] == '(' || a[i] == '{' || a[i] == '[')
        s.push(a[i]);
      else {
        if (a[i] == '>') {
          if (s.empty()) {
            flag = false;
            break;
          } else if (s.top() == '<')
            s.pop();
          else if (s.top() == '{' || s.top() == '(' || s.top() == '[') {
            k++;
            s.pop();
          }
        }
        if (a[i] == '}') {
          if (s.empty()) {
            flag = false;
            break;
          } else if (s.top() == '{')
            s.pop();
          else if (s.top() == '<' || s.top() == '(' || s.top() == '[') {
            k++;
            s.pop();
          }
        }
        if (a[i] == ']') {
          if (s.empty()) {
            flag = false;
            break;
          } else if (s.top() == '[')
            s.pop();
          else if (s.top() == '{' || s.top() == '(' || s.top() == '<') {
            k++;
            s.pop();
          }
        }
        if (a[i] == ')') {
          if (s.empty()) {
            flag = false;
            break;
          } else if (s.top() == '(')
            s.pop();
          else if (s.top() == '{' || s.top() == '<' || s.top() == '[') {
            k++;
            s.pop();
          }
        }
      }
    }
    if (!s.empty() || !flag)
      printf("Impossible\n");
    else
      printf("%d\n", k);
  }
  return 0;
}