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.
def probability(r,s,p):
    q = (r+s+p)*(r+s+p-1)/2 - r*(r-1)/2 - s*(s-1)/2 - p*(p-1)/2
    return {"rs":r*s/q, "rp":r*p/q, "sp":s*p/q}#, "rr":r*(r-1)/2/q, "ss":s*(s-1)/2/q, "pp":p*(p-1)/2/q}

def mult(array, n):
    return [a*n for a in array]

def sum(array1, array2):
    return [array1[i]+array2[i] for i in range(len(array1))]

def calculate(R,S,P):
    d = [[[[0,0,0] for p in range(P+1)]for s in range(S+1)]for r in range(R+1)]
    
    for r in range(0,R+1):
        for s in range(0,S+1):
            for p in range(0,P+1):
                if r*s + r*p + s*p == 0: #has two zeroes
                    rsp = r+s+p
                    if rsp > 0:
                        d[r][s][p] = [r/rsp,s/rsp,p/rsp]
                else:                    
                    prob = probability(r,s,p)
                    if r+s+p == 1:
                        d[r][s][p] = [r,s,p]
                    if r > 0:
                        d[r][s][p] = sum(d[r][s][p], mult(d[r-1][s][p], prob["rp"]))
                    if s > 0:
                        d[r][s][p] = sum(d[r][s][p], mult(d[r][s-1][p], prob["rs"]))
                    if p > 0:
                        d[r][s][p] = sum(d[r][s][p], mult(d[r][s][p-1], prob["sp"]))
    print(' '.join(str(n) for n in d[R][S][P]))

R,S,P = map(int, input().split())
calculate(R,S,P)