Problem: 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.

Input

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.

Output

Print three space-separated real numbers: the probabilities, at which the rocks, the scissors and the paper will be the only surviving species, respectively. The answer will be considered correct if the relative or absolute error of each number doesn't exceed 10 - 9.

Examples

Input

2 2 2


Output

0.333333333333 0.333333333333 0.333333333333


Input

2 1 2


Output

0.150000000000 0.300000000000 0.550000000000


Input

1 1 3


Output

0.057142857143 0.657142857143 0.285714285714

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int r, s, p;
int cn2(int n) {
  if (n < 2) return 0;
  return (n * (n - 1)) / 2;
}
struct node {
  double ans1;
  double ans2;
  double ans3;
} f[110][110][110];
node dp(int a, int b, int c) {
  if (f[a][b][c].ans1 != -1 && f[a][b][c].ans2 != -1 && f[a][b][c].ans3 != -1)
    return f[a][b][c];
  if (a > 0 && b > 0 && c > 0) {
    int s = cn2(a + b + c) - cn2(a) - cn2(b) - cn2(c);
    double x1 = (double)(a * b) / (double)s;
    double x2 = (double)(a * c) / (double)s;
    double x3 = (double)(b * c) / (double)s;
    double t11 = dp(a, b - 1, c).ans1, t12 = dp(a, b - 1, c).ans2,
           t13 = dp(a, b - 1, c).ans3;
    double t21 = dp(a - 1, b, c).ans1, t22 = dp(a - 1, b, c).ans2,
           t23 = dp(a - 1, b, c).ans3;
    double t31 = dp(a, b, c - 1).ans1, t32 = dp(a, b, c - 1).ans2,
           t33 = dp(a, b, c - 1).ans3;
    node txt;
    txt.ans1 = x1 * t11 + x2 * t21 + x3 * t31;
    txt.ans2 = x1 * t12 + x2 * t22 + x3 * t32;
    txt.ans3 = x1 * t13 + x2 * t23 + x3 * t33;
    return f[a][b][c] = txt;
  }
  node txt;
  int cnt = 0;
  if (a == 0) cnt++;
  if (b == 0) cnt++;
  if (c == 0) cnt++;
  if (cnt == 3) {
    txt.ans1 = 0;
    txt.ans2 = 0;
    txt.ans3 = 0;
    return f[a][b][c] = txt;
  }
  if (cnt == 2) {
    if (a == 0) txt.ans1 = 0;
    if (b == 0) txt.ans2 = 0;
    if (c == 0) txt.ans3 = 0;
    if (txt.ans1 == 0 && txt.ans2 == 0) txt.ans3 = 1;
    if (txt.ans1 == 0 && txt.ans3 == 0) txt.ans2 = 1;
    if (txt.ans2 == 0 && txt.ans3 == 0) txt.ans1 = 1;
    return f[a][b][c] = txt;
  }
  if (cnt == 1) {
    if (a == 0) {
      txt.ans1 = 0;
      txt.ans2 = 1;
      txt.ans3 = 0;
    }
    if (b == 0) {
      txt.ans1 = 0;
      txt.ans2 = 0;
      txt.ans3 = 1;
    }
    if (c == 0) {
      txt.ans1 = 1;
      txt.ans2 = 0;
      txt.ans3 = 0;
    }
    return f[a][b][c] = txt;
  }
}
int main() {
  scanf("%d%d%d", &r, &s, &p);
  for (int i = 0; i <= r; i++) {
    for (int j = 0; j <= s; j++) {
      for (int k = 0; k <= p; k++) {
        node txt;
        txt.ans1 = -1;
        txt.ans2 = -1;
        txt.ans3 = -1;
        f[i][j][k] = txt;
      }
    }
  }
  printf("%.12lf %.12lf %.12lf\n", dp(r, s, p).ans1, dp(r, s, p).ans2,
         dp(r, s, p).ans3);
  return 0;
}