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.
r, s, p = [int(x) for x in input().split()]
memo = [[[None for p_ind in range(p + 1)] for s_ind in range(s + 1)] for r_ind in range(r + 1)]      

def compute_probs(r: int, s: int, p: int):
    if memo[r][s][p] is not None:
        return memo[r][s][p]
        
    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]

    

    total = r*s + r*p + s*p
    r1_mult, r2_mult, r3_mult = r*s/total, r*p/total, s*p/total

    res1 = [r1_mult * elem for elem in compute_probs(r, s - 1, p)]
    res2 = [r2_mult * elem for elem in compute_probs(r - 1, s, p)]
    res3 = [r3_mult * elem for elem in compute_probs(r, s, p - 1)]

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

    return ans


x, y, z = compute_probs(r, s, p)



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


# t2 = datetime.datetime.now()

# print(t2 - t1)