#define MOD_TYPE 1 #pragma region Macros #include using namespace std; #if 0 #include #include using Int = boost::multiprecision::cpp_int; using lld = boost::multiprecision::cpp_dec_float_100; #endif #if 1 #pragma GCC target("avx2") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #endif using ll = long long int; using ld = long double; using pii = pair; using pll = pair; using pld = pair; template using smaller_queue = priority_queue, greater>; constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353); constexpr int INF = (int)1e9 + 10; constexpr ll LINF = (ll)4e18; constexpr ld PI = acos(-1.0); constexpr ld EPS = 1e-9; constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0}; constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0}; #define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i) #define rep(i, n) REP(i, 0, n) #define REPI(i, m, n) for (int i = m; i < (int)(n); ++i) #define repi(i, n) REPI(i, 0, n) #define MP make_pair #define MT make_tuple #define YES(n) cout << ((n) ? "YES" : "NO") << "\n" #define Yes(n) cout << ((n) ? "Yes" : "No") << "\n" #define possible(n) cout << ((n) ? "possible" : "impossible") << "\n" #define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n" #define all(v) v.begin(), v.end() #define NP(v) next_permutation(all(v)) #define dbg(x) cerr << #x << ":" << x << "\n"; struct io_init { io_init() { cin.tie(0); ios::sync_with_stdio(false); cout << setprecision(30) << setiosflags(ios::fixed); }; } io_init; template inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; } template inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; } inline ll CEIL(ll a, ll b) { return (a + b - 1) / b; } template inline void Fill(A (&array)[N], const T &val) { fill((T *)array, (T *)(array + N), val); } template constexpr istream &operator>>(istream &is, pair &p) noexcept { is >> p.first >> p.second; return is; } template constexpr ostream &operator<<(ostream &os, pair &p) noexcept { os << p.first << " " << p.second; return os; } #pragma endregion // -------------------------------------- int dp[100010]; bool f(int n) { if (n == 0) return true; if (dp[n] != -1) return dp[n]; int p = 1; bool res = false; while (p <= n) { if (!f(n - p)) { res = true; break; } p *= 3; } return dp[n] = res; } void solve() { /* Fill(dp, -1); rep(i, 100) { cout << i << " : " << f(i) << "\n"; } */ ll n; cin >> n; YES(n % 2 == 0); } int main() { solve(); }