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 PYTHON):

Solution:
# -*- coding: utf-8 -*-
import sys
import heapq
from collections import defaultdict
from collections import OrderedDict
import copy
from collections import deque


class Solver(object):

    def run(self):
        r, s, p = readarray(int)

        dp = [[[0.0]*101 for row in xrange(101)] for col in xrange(101)]
        dp[r][s][p] = 1.0

        for i in sorted(xrange(101), reverse=True):
            if r < i:
                continue
            for j in sorted(xrange(101), reverse=True):
                if s < j:
                    continue
                for k in sorted(xrange(101), reverse=True):
                    if p < k:
                        continue

                    TOTAL = i*j + j*k + k*i
                    if dp[i][j][k] != 0.0 and TOTAL != 0:
                        # r kills s
                        dp[i][j-1][k] += float(i*j)/TOTAL * dp[i][j][k]

                        # s kills p
                        dp[i][j][k-1] += float(j*k)/TOTAL * dp[i][j][k]

                        # p kills r
                        dp[i-1][j][k] += float(k*i)/TOTAL * dp[i][j][k]

        rp = 0.0
        sp = 0.0
        pp = 0.0

        for i in xrange(101):
            if dp[i][0][0] != 0.0:
                rp += dp[i][0][0]
        for i in xrange(101):
            if dp[0][i][0] != 0.0:
                sp += dp[0][i][0]
        for i in xrange(101):
            if dp[0][0][i] != 0.0:
                pp += dp[0][0][i]

        print("%2.12f %2.12f %2.12f" % (rp, sp, pp))



####################################
# ここからは見ちゃダメ！

def read(foo):
    return foo(raw_input())
def readarray(foo):
    return [foo(x) for x in raw_input().split()]
def dbg(a):
    sys.stderr.write(str(a))
def dbgln(a):
    sys.stderr.write(str(a) + "\n")
# https://gist.github.com/kachayev/5910538
def topological(graph):
    GRAY, BLACK = 0, 1
    order, enter, state = deque(), set(graph), {}
    def dfs(node):
        state[node] = GRAY
        for k in graph.get(node, ()):
            sk = state.get(k, None)
            if sk == GRAY: raise ValueError("cycle")
            if sk == BLACK: continue
            enter.discard(k)
            dfs(k)
        order.appendleft(node)
        state[node] = BLACK
    while enter: dfs(enter.pop())
    return order
# l1 to l2 -> line
# ls1 to ls2 -> line segment
def lineIntersect( l1, l2, ls1, ls2 ):
    l1x, l1y = l1
    l2x, l2y = l2
    ls1x, ls1y = ls1
    ls2x, ls2y = ls2
    if (((l1x - l2x) * (ls1y - l1y) + (l1y - l2y) * (l1x - ls1x)) *
        ((l1x - l2x) * (ls2y - l1y) + (l1y - l2y) * (l1x - ls2x)) > 0):
        return False
    return True
# p1 to p2 -> line segment
# p3 to p4 -> line segemnt
def intersect( p1, p2, p3, p4 ):
    if lineIntersect(p1, p2, p3, p4) and lineIntersect(p3, p4, p1, p2):
        return True
    return False

if __name__ == '__main__':
    Solver().run()