#include #include using namespace std; using namespace atcoder; using ll = long long; // -2^63 ~ 2^63-1 (9.2*10^18) using ull = unsigned long long; // 0 ~ 2^64-1 (1.8*10^19) using ld = long double; #define rep(i, n) for (int i = 0; i < (int)(n); i++) #define rep2(i, m, n) for (int i = (m); (i) < int (n); i++) #define all(v) v.begin(), v.end() #define bit(n) (1ll<<(n)) // 2^n #define sz(x) ((int)(x).size()) #define fi first #define se second template using maxpq = priority_queue; template using minpq = priority_queue, greater>; templatebool chmax(T &a, const T &b) { if (abool chmin(T &a, const T &b) { if (b auto make_vec(const ll (&d)[n], T&& init) noexcept { if constexpr (idx < n) return std::vector(d[idx], make_vec(d, std::forward(init))); else return init; } template auto make_vec(const ll (&d)[n], const T& init = {}) noexcept { if constexpr (idx < n) return std::vector(d[idx], make_vec(d, init)); else return init; } // auto 変数名 = make_vec<型>({a, b, ...}, 初期値); int main(){ ll N, K; cin >> N >> K; vector a(N); rep(i, N){ cin >> a[i]; } ll ans = 0; function&)> dfs = [&](vector &A){ // 数列の長さが N に達したら打ち切り if (A.size() == K) { // 処理 ll S = 0; rep(i, K){ S += a[A[i]]; } //cout << S << endl; if(S%998244353 <= S%998) ans++; return; } // rep2(v, Aの最後の要素, M)とすればAが狭義単調増加などの場合も対応できる // Aが空の場合に注意 int prev_last = (A.empty() ? 0 : A.back()+1); rep2(v, prev_last, N){ A.push_back(v); dfs(A); A.pop_back(); } }; vector A; dfs(A); cout << ans%998 << endl; }