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>
using namespace std;
const int maxn = 1000005;
char s[maxn];
int main() {
  while (~scanf("%s", s)) {
    stack<char> st;
    int len = strlen(s);
    bool flag = true;
    int ans = 0;
    for (int i = 0; i < len; i++) {
      if (s[i] == '(' || s[i] == '[' || s[i] == '<' || s[i] == '{')
        st.push(s[i]);
      else {
        if (!st.empty()) {
          char tmp = st.top();
          st.pop();
          if (tmp == '(' && s[i] != ')') ans++;
          if (tmp == '[' && s[i] != ']') ans++;
          if (tmp == '<' && s[i] != '>') ans++;
          if (tmp == '{' && s[i] != '}') ans++;
        } else {
          flag = false;
        }
      }
    }
    if (!flag || !st.empty())
      printf("Impossible\n");
    else
      printf("%d\n", ans);
  }
  return 0;
}
