Problem: The Bad Luck Island is inhabited by three kinds of species: r rocks, s scissors and p papers. At some moments of time two random individuals meet (all pairs of individuals can meet equiprobably), and if they belong to different species, then one individual kills the other one: a rock kills scissors, scissors kill paper, and paper kills a rock. Your task is to determine for each species what is the probability that this species will be the only one to inhabit this island after a long enough period of time.

Input

The single line contains three integers r, s and p (1 ≤ r, s, p ≤ 100) — the original number of individuals in the species of rock, scissors and paper, respectively.

Output

Print three space-separated real numbers: the probabilities, at which the rocks, the scissors and the paper will be the only surviving species, respectively. The answer will be considered correct if the relative or absolute error of each number doesn't exceed 10 - 9.

Examples

Input

2 2 2


Output

0.333333333333 0.333333333333 0.333333333333


Input

2 1 2


Output

0.150000000000 0.300000000000 0.550000000000


Input

1 1 3


Output

0.057142857143 0.657142857143 0.285714285714

Solution (in JAVA):

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

public class Main {
    static FastScanner in;
    static PrintWriter out;
    static final int MOD = 1000000007;

    public static void main(String[] args) throws IOException {
        //        Scanner in = new Scanner(new File("input.txt"));
        //        Scanner in = new Scanner(System.in);
        //        System.setOut(new PrintStream(new BufferedOutputStream(new FileOutputStream("output.txt")), true));
        out = new PrintWriter(System.out);
        in = new FastScanner(System.in);
        //        in = new FastScanner("input.txt");
        int r = in.nextInt();
        int s = in.nextInt();
        int p = in.nextInt();
        double dp[][][] = new double[r + 2][s + 2][p + 2];
        dp[r][s][p] = 1;
        Queue<State> q = new LinkedList<State>();
        boolean isVisited[][][] = new boolean[r + 2][s + 2][p + 2];
        q.add(new State(r, s, p));
        isVisited[r][s][p] = true;
        while (!q.isEmpty()) {
            State ss = q.poll();
            int i = ss.i;
            int j = ss.j;
            int k = ss.k;
            double prob = cal(k + 1, j, i);
            dp[i][j][k] += prob * dp[i][j][k + 1];

            prob = cal(i, j + 1, k);
            dp[i][j][k] += prob * dp[i][j + 1][k];

            prob = cal(i + 1, k, j);
            dp[i][j][k] += prob * dp[i + 1][j][k];

            if (i > 0) {
                if (!isVisited[i - 1][j][k]) {
                    q.add(new State(i - 1, j, k));
                    isVisited[i - 1][j][k] = true;
                }
            }
            if (j > 0) {
                if (!isVisited[i][j - 1][k]) {
                    q.add(new State(i, j - 1, k));
                    isVisited[i][j - 1][k] = true;
                }
            }
            if (k > 0) {
                if (!isVisited[i][j][k - 1]) {
                    q.add(new State(i, j, k - 1));
                    isVisited[i][j][k - 1] = true;
                }
            }
        }
        double rr = 0, ss = 0, pp = 0;
        for (int i = 1; i <= r; i++) {
            rr += dp[i][0][0];
        }
        for (int i = 1; i <= s; i++) {
            ss += dp[0][i][0];
        }
        for (int i = 1; i <= p; i++) {
            pp += dp[0][0][i];
        }
        out.printf("%.9f ", rr);
        out.printf("%.9f ", ss);
        out.printf("%.9f ", pp);
        out.close();
    }

    static double cal(int a, int b, int c) {
        if (a == 0 && b == 0 || (b == 0 && c == 0) || (a == 0 && c == 0))
            return 0;
        return (double) (a * b) / (a * b + b * c + c * a);
    }
}

class State {
    public int i;
    public int j;
    public int k;

    public State(int i, int j, int k) {
        this.i = i;
        this.j = j;
        this.k = k;
    }
}

class Pair {
    public int first; //first member of pair
    public int second; //second member of pair

    public Pair(int first, int second) {
        this.first = first;
        this.second = second;
    }

    public Pair() {

    }
}

class FastScanner {

    BufferedReader br;
    StringTokenizer tokenizer;

    FastScanner(String fileName) throws FileNotFoundException {
        this(new FileInputStream(new File(fileName)));
    }

    FastScanner(InputStream is) {
        br = new BufferedReader(new InputStreamReader(is));
    }

    String nextLine() throws IOException {
        tokenizer = null;
        return br.readLine();
    }

    String next() throws IOException {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            String line = br.readLine();
            if (line == null) {
                return null;
            }
            tokenizer = new StringTokenizer(line);
        }
        return tokenizer.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    char nextChar() throws IOException {
        return next().charAt(0);
    }
}