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.
# aadiupadhyay
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")


if os.path.exists('input.txt'):
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')


A, B, C = mp()
A += 1
B += 1
C += 1
M = max(A, B, C)
p = [[[0] * (M) for i in range(M)] for j in range(M)]

for a in range(M):
    for b in range(M):
        for c in range(M):
            val = 0
            if a == 0 or b == 0:
                val = 0
            elif c == 0:
                val = 1
            else:
                div = a*b + b*c + c*a
                val = (a*b) / div * p[a][b-1][c] + (b*c) / \
                    div * p[a][b][c-1] + (a*c) / div * p[a-1][b][c]

            p[a][b][c] = val

print(p[A-1][B-1][C-1], p[B-1][C-1][A-1], p[C-1][A-1][B-1])
