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
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const long long MOD = (int)1e9 + 7;
string ascii_lowercase = "abcdefghijklmnopqrstuvwxyz";
string ascii_uppercase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
struct hashLL {
  size_t operator()(long long x) const {
    x = (x ^ (x >> 30)) * UINT64_C(0xbf58476d1ce4e5b9);
    x = (x ^ (x >> 27)) * UINT64_C(0x94d049bb133111eb);
    x = x ^ (x >> 31);
    return x;
  }
};
long long inv(long long a, long long b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
template <class T>
void coutend(T arg) {
  cout << arg << '\n';
}
template <class T>
void coutsp(T arg) {
  cout << arg << ' ';
}
template <class T>
void coutarrsp(T arr, int l, int r) {
  for (int i = l; i <= r; i++) {
    cout << arr[i] << ' ';
  }
  cout << '\n';
}
template <class T>
void coutarr(T arr, int l, int r) {
  for (int i = l; i <= r; i++) {
    cout << arr[i];
  }
  cout << '\n';
}
const long long INF = 1e18;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long distsq(long long x1, long long y1, long long x2, long long y2) {
  return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
}
const int N = 101;
long long n, a, b, t, q, k, m, r, s, p;
double dp[N][N][N][3];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> r >> s >> p;
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < (r + 1); i++) {
    for (int j = 0; j < (p + 1); j++) {
      for (int k = 0; k < (s + 1); k++) {
        if (i + j + k == 0) continue;
        if (i == 0 && j == 0)
          dp[i][j][k][2] = 1.0;
        else if (i == 0 && k == 0)
          dp[i][j][k][1] = 1.0;
        else if (j == 0 && k == 0)
          dp[i][j][k][0] = 1.0;
        else {
          double rp = i * j;
          double rs = i * k;
          double ps = j * k;
          double tot = rp + rs + ps;
          if (rp > 0) {
            for (int z = 0; z < (3); z++) {
              dp[i][j][k][z] += (rp / tot) * dp[i - 1][j][k][z];
            }
          }
          if (rs > 0) {
            for (int z = 0; z < (3); z++) {
              dp[i][j][k][z] += (rs / tot) * dp[i][j][k - 1][z];
            }
          }
          if (ps > 0) {
            for (int z = 0; z < (3); z++) {
              dp[i][j][k][z] += (ps / tot) * dp[i][j - 1][k][z];
            }
          }
        }
      }
    }
  }
  cout << fixed << setprecision(15);
  coutsp(dp[r][p][s][0]);
  coutsp(dp[r][p][s][2]);
  coutsp(dp[r][p][s][1]);
  cout << '\n';
}
