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.
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.
import sys
input = sys.stdin.readline
def I():return input().strip()
def II():return int(input().strip())
def LI():return [*map(int,input().strip().split())]
import string, math, time, functools, random, fractions
from heapq import heappush, heappop, heapify
from bisect import bisect_left, bisect_right
from collections import deque, defaultdict, Counter, OrderedDict
from itertools import permutations, combinations, groupby
from operator import itemgetter

rock, scissor, paper = LI()
dp = [[[0 for p in range(paper + 1)] for s in range(scissor + 1)] for r in range(rock + 1)]
dp[rock][scissor][paper] = 1
for r in range(rock, -1, -1):
    for s in range(scissor, -1, -1):
        for p in range(paper, -1, -1):
            currsum = p * r + r * s + s * p
            if not currsum:
                continue
            if r >= 1:
                dp[r - 1][s][p] += (p * r) / currsum * dp[r][s][p]
            if s >= 1:
                dp[r][s - 1][p] += (r * s) / currsum * dp[r][s][p]
            if p >= 1:
                dp[r][s][p - 1] += (s * p) / currsum * dp[r][s][p]
pr = sum(dp[i][0][0] for i in range(1, rock + 1))
ps = sum(dp[0][i][0] for i in range(1, scissor + 1))
pp = sum(dp[0][0][i] for i in range(1, paper + 1))
print(round(pr, 12), round(ps, 12), round(pp, 12))