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.
from decimal import getcontext, Decimal
from typing import List
from functools import cache
import datetime


class Solution:
    def __init__(self):
        self.memo = [[[[float('inf'), float('inf'), float('inf')] for p_ind in range(p + 1)] for s_ind in range(s + 1)] for r_ind in range(r + 1)]

    def compute_probs(self, r: int, s: int, p: int) -> List[int]:
        if r == 0:
            return [0, 1, 0]
        if s == 0:
            return [0, 0, 1]
        if p == 0:
            return [1, 0, 0]
        if r == s == p:
            return [1/3, 1/3, 1/3]

        if self.memo[r][s][p] != [float('inf'), float('inf'), float('inf')]:
            return self.memo[r][s][p]

        total = r*s + r*p + s*p

        res1 = [r*s/total * elem for elem in self.compute_probs(r, s - 1, p)]
        res2 = [r*p/total * elem for elem in self.compute_probs(r - 1, s, p)]
        res3 = [s*p/total * elem for elem in self.compute_probs(r, s, p - 1)]

        ans = [x + y + z for x, y, z in zip(res1, res2, res3)]
        self.memo[r][s][p] = ans

        return ans

    

r, s, p = [int(x) for x in input().split()]
solution_obj = Solution()
x, y, z = solution_obj.compute_probs(r, s, p)



print(f"{str(x)} {str(y)} {str(z)}")


# t2 = datetime.datetime.now()

# print(t2 - t1)