Problem: IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.

The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.

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Input

The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.

Output

Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.

Examples

Input

2 5 3


Output

38.546168065709

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void init() {
  ios::sync_with_stdio(0);
  srand(time(0));
}
void pause() { puts(""); }
void scanBigInteger(vector<short> &A) {
  string s;
  cin >> s;
  for (int x = s.length() - 1; x >= 0; x--) {
    A.push_back(s[x] - '0');
  }
}
void printBigInteger(vector<short> A) {
  if (A.empty())
    cout << 0 << endl;
  else
    for (int x = A.size() - 1; x >= 0; x--) cout << A[x];
}
void swap(vector<short> &A, vector<short> &B) {
  vector<short> C = A;
  A = B;
  B = C;
}
void convert(vector<short> &A, vector<short> &B) {
  if (A.size() > B.size())
    while (abs((int)A.size() - (int)B.size())) B.push_back(0);
  if (A.size() < B.size())
    while (abs((int)A.size() - (int)B.size())) A.push_back(0);
}
void normalize(vector<short> &A) {
  while (A.size() && A[A.size() - 1] == 0) A.pop_back();
}
bool operator>(vector<short> A, vector<short> B) {
  if (A.size() > B.size()) return true;
  if (A.size() < B.size())
    return false;
  else {
    for (int i = A.size() - 1; i >= 0; i--) {
      if (A[i] == B[i]) continue;
      if (A[i] > B[i]) return true;
      if (A[i] < B[i]) return false;
    }
    return false;
  }
}
bool operator<(vector<short> A, vector<short> B) {
  if (A.size() > B.size()) return false;
  if (A.size() < B.size())
    return true;
  else {
    for (int i = A.size() - 1; i >= 0; i--) {
      if (A[i] == B[i]) continue;
      if (A[i] > B[i]) return false;
      if (A[i] < B[i]) return true;
    }
    return false;
  }
}
bool operator!=(vector<short> A, vector<short> B) {
  if (A.size() != B.size()) return true;
  bool notequal = false;
  for (int i = 0; i < A.size() && !notequal; i++) {
    if (A[i] != B[i]) notequal = true;
  }
  return notequal;
}
bool operator==(vector<short> A, vector<short> B) {
  if (A.size() != B.size()) return false;
  bool equal = true;
  for (int i = 0; i < A.size() && equal; i++) {
    if (A[i] != B[i]) equal = false;
  }
  return equal;
}
bool operator<=(vector<short> A, vector<short> B) { return (A < B) | (A == B); }
bool operator>=(vector<short> A, vector<short> B) { return (A > B) | (A == B); }
vector<short> operator+(vector<short> A, vector<short> B) {
  vector<short> C;
  convert(A, B);
  int carry = 0;
  for (int i = 0; i < A.size(); i++) {
    carry += A[i] + B[i];
    C.push_back(carry % 10);
    carry /= 10;
  }
  if (carry) C.push_back(carry);
  return C;
}
vector<short> operator-(vector<short> A, vector<short> B) {
  if (B > A) swap(A, B);
  int carry = 0;
  for (int i = 0; i < A.size(); i++) {
    A[i] -= carry + int(i < B.size() ? B[i] : 0);
    carry = int(A[i] < 0);
    if (carry) A[i] += 10;
  }
  while (A.size() && A[A.size() - 1] == 0) A.pop_back();
  return A;
}
vector<short> operator*(vector<short> A, vector<short> B) {
  if (B > A) swap(A, B);
  vector<short> C;
  C.resize(A.size() + B.size());
  int slice = 0;
  for (int i = 0; i < A.size(); i++) {
    for (int j = 0; j < B.size() || slice; j++) {
      int tmp = C[i + j] + slice + (j < B.size() ? B[j] : 0) * A[i];
      C[i + j] = tmp % 10;
      slice = tmp / 10;
    }
  }
  while (C.size() && C[C.size() - 1] == 0) C.pop_back();
  return C;
}
long long power(long double a, long long b) {
  if (b < 2) return b ? a : 1;
  return (power((a * a), b / 2) * (b % 2 ? a : 1));
}
static long double pi = acos(-1);
long double R(long double a, long double n) {
  return (a) / (2.0 * sin(pi / n));
}
long double H(long double r, long double l) { return sqrtf(l * l - r * r); }
long double V(long double h, long double l, int n) {
  return (n * l * l * h) / (12 * tan(pi / n));
}
int main() {
  init();
  long double l1, l2, l3;
  cin >> l1 >> l2 >> l3;
  long double ans = 0;
  ans = (l1 * l1 * l1 * sqrt(2)) / 12;
  ans += l2 * l2 * l2 / (3 * sqrt(2));
  ans += l3 * l3 * l3 * sqrt((5 + 2 * sqrt(5)) * (10 - 2 * sqrt(5))) / 24;
  cout << fixed << setprecision(10) << ans << endl;
  pause();
  return 0;
}