Problem:
Score: 100 points

Problem Statement

The weather in Takahashi's town changes day by day, in the following cycle: Sunny, Cloudy, Rainy, Sunny, Cloudy, Rainy, ...

Given is a string S representing the weather in the town today. Predict the weather tomorrow.

Constraints

S is Sunny, Cloudy, or Rainy.

Input

Input is given from Standard Input in the following format:

S

Output

Print a string representing the expected weather tomorrow, in the same format in which input is given.

Sample Input 1

Sunny

Sample Output 1

Cloudy

In Takahashi's town, a sunny day is followed by a cloudy day.

Sample Input 2

Rainy

Sample Output 2

Sunny

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <iomanip>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
//typedef pair<ll, ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)
#define pb push_back
#define f first
#define s second
#define chmin(x, y) x = max(x, y);
/*
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll max3(ll a, ll b, ll c){return max(a, max(b, c));};
ll min3(ll a, ll b, ll c){return min(a, min(b, c));};
ll max4(ll a, ll b, ll c, ll d){return max(max(a, b), min(c, d));};
ll min4(ll a, ll b, ll c, ll d){return min(min(a, b), min(c, d));};
ll max5(ll a, ll b, ll c, ll d, ll e){return max(max(a, b), max3(c, d, e));};
ll min5(ll a, ll b, ll c, ll d, ll e){return min(min(a, b), min3(c, d, e));};
*/
const ll INFL = 1LL << 60;//10^18 = 2^60
//const int INF = 1 << 30;//10^9
const ll INF = INFL;
//ll MOD = 1000000007;
ll MOD = 998244353;

vector<ll> dy = {0, 0, 1, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, -1, 0, 0, 1, -1, 1, -1, 0};

//デバッグ
void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}
ll get_bit(ll x){return __builtin_popcountll(x);}

ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}

ll pow_mod(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2){
      res *= x; res %= MOD;
    }
    x *= x; x %= MOD;
    k /= 2;
  }
  return res;
}

ll inverse(ll x){return pow_mod(x, MOD - 2);};

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y){return x / gcd(x, y) * y;};

ll kai_mod(ll x){
  if(x == 0) return 1;
  return x * kai_mod(x-1) % MOD;
}

//約数の列挙O(√n)
vector<ll> divisor(ll n){
    vector<ll> res(0);
    for(ll i = 1; i * i <= n; i++){
        if(n % i == 0){
            res.push_back(i);
            if(i != n/i) res.push_back(n/i);
        }
    }
    sort(res.begin(), res.end());
    return res;
}

//素因数列挙
vector<ll> MakePrimeList(ll x) {
    vector<ll> res;
    for(ll k=2; k*k<=x; k++) {
        if(x % k == 0) {
            res.pb(k);
            while(x % k == 0) x /= k;
        }
    }
    if(x > 1) res.push_back(x);
    return res;
}


//コンビネーション
const int MAXcomb = 200010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i = 2; i < MAXcomb; i++){
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
ll comb(int n, int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
}

int main(){
  string S;
  cin >> S;
  if(S == "Sunny") cout << "Cloudy" << endl;
  else if(S == "Cloudy") cout << "Rainy" << endl;
  else cout << "Sunny" << endl;
}