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.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Stack;


public class CEdu4 {
	public static void main(String[] args) throws IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		char[] o = {'(', '<', '[', '{'}, e = {')', '>', ']', '}'};
//		for(int i = 0; i < 4; i++) System.out.println(o[i]);
		long osum = 0, esum = 0;
		String line = f.readLine();
		for(int i = 0; i < line.length(); i++) {
			if(Arrays.binarySearch(o, line.charAt(i)) >= 0) osum++;
			else esum++;
		}
		if(osum != esum) {
			System.out.println("Impossible");
			return;
		}
		Stack<Integer> stack = new Stack<Integer>();
		long ans = 0;
		for(int i = 0; i < line.length(); i++) {
			if(line.charAt(i) == '(') stack.push(0);
			else if(line.charAt(i) == '<') stack.push(1);
			else if(line.charAt(i) == '[') stack.push(2);
			else if(line.charAt(i) == '{') stack.push(3);
			else {
				if(stack.isEmpty()) {
					System.out.println("Impossible");
					return;
				}
				int cur = stack.peek();
				char a = (char) line.charAt(i);
//				System.out.println(a);
				int ind = Arrays.binarySearch(e, a);
//				System.out.println(cur);
//				System.out.println(ind);
//				System.out.println();
				if(cur == ind) stack.pop();
				else {
					stack.pop();
					ans++;
				}
			}
//			System.out.println(stack.toString());
//			System.out.println("CurAns: " + ans);
		}
		System.out.println(ans);
	}
}

// (([{>}{[{[)]]>>]
// 