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 JAVA):

Solution:
import java.io.*;
import java.util.Stack;
import java.util.StringTokenizer;


public class ProblemA {
    public static void main(String[] args) {
        int testMode = 0;
        String testInput = "]]";
        /*-------------------------------------------------------------------------------------------------------*/
        FastReader in = new FastReader(testMode == 0 ? System.in : new ByteArrayInputStream(testInput.getBytes()));
        PrintWriter out = new PrintWriter(System.out);
        new ProblemSolver().solve(in, out);
        out.close();
    }
}

class ProblemSolver {
    int N;

    boolean f(char c) {
        return (c == '[' || c == '(' || c == '{' || c == '<');
    }

    char g(char c) {
        if (c == '}') return '{';
        if (c == ']') return '[';
        if (c == '>') return '<';
        return '(';
    }

    void solve(FastReader in, PrintWriter out) {
        char A[] = in.nextToken().toCharArray();
        Stack<Character> S = new Stack<Character>();
        if (A.length == 0) {
            out.println(0);
            return;
        }
        if (!f(A[0])) {
            out.println("Impossible");
            return;
        }
        int res = 0;
        S.add(A[0]);
        for (int i = 1; i < A.length; i++) {
            if (f(A[i])) {
                S.add(A[i]);
            } else {
                if (S.size() == 0) {
                    out.println("Impossible");
                    return;
                }
                res += (S.peek() == g(A[i]) ? 0 : 1);
                S.pop();
            }
        }
        if (S.size() > 0) {
            out.println("Impossible");
            return;
        }

        out.println(res);
    }
}

class Pair implements Comparable<Pair> {
    public int a;
    public int b;

    public Pair(int a, int b) {
        this.a = a;
        this.b = b;
    }

    public int compareTo(Pair o) {
        if (this.a < o.a) return -1;
        if (this.a > o.a) return 1;
        if (this.b == o.b) return 0;
        return (int) ((this.b - o.b) / Math.abs(this.b - o.b));
    }
}


class FastReader extends BufferedReader {
    private StringTokenizer st;

    public FastReader(InputStream is) {
        super(new InputStreamReader(is));
    }

    public String nextToken() {
        try {
            while (st == null || !st.hasMoreElements()) {
                st = new StringTokenizer(readLine());
            }
            return st.nextToken();
        } catch (IOException e) {
            return null;
        }
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

    public int[] nextArrayInt(int size) {
        int[] ret = new int[size];
        for (int i = 0; i < size; i++) {
            ret[i] = nextInt();
        }
        return ret;
    }
}