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
import java.io.BufferedReader;
import java.io.InputStreamReader;


public class ReplaceToMakeRegularBracketSequence {

	/**
	 * @param args
	 */
	public static void main(String[] args)throws Exception {
		// TODO Auto-generated method stub

				BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
				char[] cc = br.readLine().toCharArray();
				int n = cc.length;
				char[] stack = new char[n];
				int k = 0, cnt = 0;
				for (int i = 0; i < n; i++) {
					char q = cc[i];
					if (q == '>' || q == '}' || q == ']' || q == ')') {
						if (k == 0) {
							System.out.println("Impossible");
							return;
						}
						char p = stack[--k];
						if ((p == '<' && q == '>' || p == '{' && q == '}' || p == '[' && q == ']' || p == '(' && q == ')'))
						{}
						else
							cnt++;
					} else
						stack[k++] = q;
				}
				System.out.println(k == 0 ? cnt : "Impossible");
			}
		}

	
