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 N = 1e6 + 6;
char s[N];
stack<int> q[4];
void no() {
  puts("Impossible");
  exit(0);
}
int main() {
  int n, m;
  gets(s);
  n = strlen(s);
  if (n & 1) no();
  string z = "()[]{}<>";
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 8; ++j)
      if (s[i] == z[j]) {
        s[i] = j + '0';
      }
  }
  for (int i = 0; i < 4; ++i) q[i].push(-1);
  for (int k = 0; k < n; ++k) {
    int x = s[k] - '0';
    if (x & 1) {
      int h = 0;
      for (int i = 0; i < 4; ++i)
        if (q[h].top() < q[i].top()) h = i;
      if (q[h].top() == -1) no();
      if (h != x / 2) ++ans;
      q[h].pop();
    } else {
      q[x / 2].push(k);
    }
  }
  for (int i = 0; i < 4; ++i)
    if (q[i].top() != -1) no();
  cout << ans << endl;
  return 0;
}
