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;
char c[1000006];
stack<int> s1;
stack<int> s2;
int main() {
  while (~scanf("%s", c)) {
    int ans = 0;
    for (int i = 0; c[i]; i++) {
      if (c[i] == '{') {
        s1.push(1);
      }
      if (c[i] == '[') {
        s1.push(2);
      }
      if (c[i] == '<') {
        s1.push(3);
      }
      if (c[i] == '(') {
        s1.push(4);
      }
      if (c[i] == '}') {
        if (s1.empty()) {
          puts("Impossible");
          return 0;
        }
        int k = s1.top();
        s1.pop();
        if (k != 1) {
          ans++;
        }
      }
      if (c[i] == ']') {
        if (s1.empty()) {
          puts("Impossible");
          return 0;
        }
        int k = s1.top();
        s1.pop();
        if (k != 2) {
          ans++;
        }
      }
      if (c[i] == ')') {
        if (s1.empty()) {
          puts("Impossible");
          return 0;
        }
        int k = s1.top();
        s1.pop();
        if (k != 4) {
          ans++;
        }
      }
      if (c[i] == '>') {
        if (s1.empty()) {
          puts("Impossible");
          return 0;
        }
        int k = s1.top();
        s1.pop();
        if (k != 3) {
          ans++;
        }
      }
    }
    if (!s1.empty()) {
      puts("Impossible");
      return 0;
    }
    printf("%d\n", ans);
  }
  return 0;
}