Problem: <image>

You have one chip and one chance to play roulette. Are you feeling lucky?

Output

Print your bet. Your chip must be placed entirely within some square (not on an edge or a corner shared by adjacent squares).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y);
long long mpower(long long x, long long y, long long p);
long long modInv(long long a, long long m);
long long gcdExtended(long long a, long long b, long long *x, long long *y);
bool isPrime(long long n);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, i, j, k, l, p, q, r, x, y, z, a, b, c, d, f, n, m,
      M = 1000000007;
  string s;
  vector<long long> v, w;
  vector<long long>::iterator itrv;
  unordered_map<long long, long long> N;
  set<long long> S;
  set<long long>::iterator itr;
  cout << "Red";
  return 0;
}
long long modInv(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  long long res = (x % m + m) % m;
  return res;
}
long long gcdExtended(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 gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long mpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  long long p = sqrt(n);
  for (int i = 5; i <= p; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}