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 int N = 1e6 + 10;
int n, a[N], sol;
string s;
stack<int> box;
int main() {
  cin >> s;
  n = s.length();
  for (int i = 0; i < n; i++) {
    if (s[i] == '[')
      a[i] = 1;
    else if (s[i] == '{')
      a[i] = 2;
    else if (s[i] == '(')
      a[i] = 3;
    else if (s[i] == '<')
      a[i] = 4;
    else if (s[i] == ']')
      a[i] = -1;
    else if (s[i] == '}')
      a[i] = -2;
    else if (s[i] == ')')
      a[i] = -3;
    else
      a[i] = -4;
  }
  if (a[0] < 0) {
    puts("Impossible");
    return 0;
  }
  box.push(a[0]);
  for (int i = 1; i < n; i++) {
    if (box.size() == 0) {
      if (a[i] > 0)
        box.push(a[i]);
      else {
        printf("Impossible");
        return 0;
      }
      continue;
    }
    int cur = box.top();
    if (a[i] > 0)
      box.push(a[i]);
    else if (cur + a[i] != 0) {
      sol++;
      box.pop();
    } else if (cur + a[i] == 0)
      box.pop();
  }
  if (box.size() > 0)
    printf("Impossible");
  else
    printf("%d", sol);
  return 0;
}