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.
# LUOGU_RID: 93944796
# -*- coding:UTF-8 -*-
# r,s,p =int(input()),int(input()),int(input())
# 一行读入3个数
r,s,p = [int(x) for x in input().split()]
# 开一个三维数组r*s*p

dp = [[[0 for i in range(p+5)] for j in range(s+5)] for k in range(r+5)]
for i in range(0,r+1):
    for j in range(0,s+1):
        for k in range(0,p+1):
            dp[i][j][k] = 0.0
dp[r][s][p] = 1.0

# for(int i = r;i >= 0;i--) {
#         for(int j = s;j >= 0;j--) {
#             for(int k = p;k >= 0;k--) {
#                 if(i == 0 && j == 0) {
#                     continue;
#                 }
#                 if(i == 0 && k == 0) {
#                     continue;
#                 }
#                 if(j == 0 && k == 0) {
#                     continue;
#                 }
#                 double sum = i*j + j*k + k*i;
#                 if(i > 0) {
#                     dp[i-1][j][k] += dp[i][j][k]*i*k/sum;
#                 }
#                 if(j > 0) {
#                     dp[i][j-1][k] += dp[i][j][k]*i*j/sum;
#                 }
#                 if(k > 0) {
#                     dp[i][j][k-1] += dp[i][j][k]*k*j/sum;
#                 }
#             }
#         }
# }
for i in range(r,-1,-1):
    for j in range(s,-1,-1):
        for k in range(p,-1,-1):
            if i == 0 and j == 0:
                continue
            if i == 0 and k == 0:
                continue
            if j == 0 and k == 0:
                continue
            sum = i*j + j*k + k*i
            if i > 0:
                dp[i-1][j][k] += dp[i][j][k]*i*k/sum
            if j > 0:
                dp[i][j-1][k] += dp[i][j][k]*i*j/sum
            if k > 0:
                dp[i][j][k-1] += dp[i][j][k]*k*j/sum

ans = 0.0
for i in range(1, r+1):
    ans = ans + dp[i][0][0]
# print(ans)
# 打印10位小数
print("%.15f" % ans)
ans = 0.0
for i in range(1, s+1):
    ans = ans + dp[0][i][0]
    # ans += dp[0][i][0]
# print(ans)
print("%.15f" % ans)
ans = 0.0
for i in range(1, p+1):
    # ans += dp[0][0][i]
    ans = ans + dp[0][0][i]
print("%.15f" % ans)