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 count(r, s, p):
	if (r == 0 and p == 0) or (r == 0 and s == 0) or (p == 0 and s == 0):
		return 0
	return r*p/(r*s+s*p+p*r)


r, s, p = map(int, input().split())
dp = [[[0] * 102 for i in range(102)] for i in range(102)]
for r1 in range(r, -1, -1):
	for s1 in range(s, -1, -1):
		for p1 in range(p, -1, -1):
			if r1 == 0 and s1 == 0 and p1 == 0:
				pass
			elif r1 == r and s1 == s and p1 == p:
				dp[r][s][p] = 10**12
			else:
				dp[r1][s1][p1] = dp[r1+1][s1][p1]*count(r1+1, s1, p1) + dp[r1][s1][p1+1]*count(p1+1, r1, s1) + dp[r1][s1+1][p1]*count(s1+1, p1, r1)
ans1, ans2, ans3 = 0, 0, 0
for i in range(1, 101):
	ans1 += dp[i][0][0]
	ans2 += dp[0][i][0]
	ans3 += dp[0][0][i]
s = ans1 + ans2 + ans3
print(ans1/s, ans2/s, ans3/s)