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;
char arr[1000000 + 10];
int main() {
  while (cin >> arr) {
    int l;
    int num = 0;
    int flag = 1;
    l = strlen(arr);
    int zuo = 0;
    int you = 0;
    stack<char> sta;
    for (int i = 0; i < l; i++) {
      if (arr[i] == '(' || arr[i] == '{' || arr[i] == '[' || arr[i] == '<') {
        sta.push(arr[i]);
        zuo++;
      } else {
        you++;
        if (!sta.empty()) {
          if (sta.top() == '(' && arr[i] == ')' ||
              sta.top() == '{' && arr[i] == '}' ||
              sta.top() == '[' && arr[i] == ']' ||
              sta.top() == '<' && arr[i] == '>')
            sta.pop();
          else {
            sta.pop();
            num++;
          }
        } else
          flag = 0;
      }
    }
    if (flag && zuo == you)
      cout << num << endl;
    else
      cout << "Impossible" << endl;
  }
  return 0;
}
