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 = 100100;
const int INF = 1E9;
char s[10 * maxn];
char st[10 * maxn];
int top = 0;
bool open(char c) {
  if (c == '{') return 1;
  if (c == '[') return 1;
  if (c == '(') return 1;
  if (c == '<') return 1;
  return 0;
}
bool eq(char c, char d) {
  if (c == '{' && d == '}') return 1;
  if (c == '(' && d == ')') return 1;
  if (c == '[' && d == ']') return 1;
  if (c == '<' && d == '>') return 1;
  return 0;
}
int solve() {
  gets(s);
  int n = strlen(s);
  int ans = 0;
  for (int i = 0; i < n; i++)
    if (open(s[i]))
      st[top++] = s[i];
    else {
      if (top == 0) {
        printf("Impossible\n");
        return 0;
      }
      top--;
      if (!eq(st[top], s[i])) ans++;
    }
  if (top != 0)
    printf("Impossible\n");
  else
    printf("%d\n", ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  solve();
  return 0;
}
