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;
string s;
int a[1500000];
vector<int> last;
int ans = 0;
int sk[5];
void func() {
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(') {
      a[i] = 1;
      continue;
    }
    if (s[i] == '[') {
      a[i] = 2;
      continue;
    }
    if (s[i] == '{') {
      a[i] = 3;
      continue;
    }
    if (s[i] == '<') {
      a[i] = 4;
      continue;
    }
    if (s[i] == ')') {
      a[i] = -1;
      continue;
    }
    if (s[i] == ']') {
      a[i] = -2;
      continue;
    }
    if (s[i] == '}') {
      a[i] = -3;
      continue;
    }
    if (s[i] == '>') {
      a[i] = -4;
      continue;
    }
  }
  return;
}
int main() {
  cin >> s;
  int sum = 0;
  func();
  for (int i = 0; i < s.size(); i++) {
    if (a[i] > 0)
      sum++;
    else
      sum--;
    if (sum < 0) {
      cout << "Impossible" << endl;
      return 0;
    }
  }
  if (sum != 0) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 0; i < s.size(); i++) {
    if (a[i] > 0) {
      sk[a[i]]++;
      last.push_back(a[i]);
    } else {
      if (sk[-a[i]] > 0 && last[last.size() - 1] == -a[i]) {
        sk[-a[i]]--;
        last.pop_back();
        continue;
      } else {
        ans++;
        last.pop_back();
      }
    }
  }
  cout << ans << endl;
}