#pragma GCC target("avx2") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #include #include // cout, endl, cin #include // string, to_string, stoi #include // vector #include // min, max, swap, sort, reverse, lower_bound, upper_bound #include // pair, make_pair #include // tuple, make_tuple #include // int64_t, int*_t #include // printf #include // map #include // queue, priority_queue #include // set #include // stack #include // deque #include // unordered_map #include // unordered_set #include // bitset #include // isupper, islower, isdigit, toupper, tolower #include #include #include using namespace std; using namespace atcoder; #define rep(i, n) for (int i = 0; i < (int)(n); i++) #define repi(i, a, b) for (int i = (int)(a); i < (int)(b); i++) typedef long long ll; typedef unsigned long long ull; const ll inf=1e18; using graph = vector > ; using P= pair; using vi=vector; using vvi=vector; using vll=vector; using vvll=vector; using vp=vector

; using vvp=vector; using vd=vector; using vvd =vector; //string T="ABCDEFGHIJKLMNOPQRSTUVWXYZ"; //string S="abcdefghijklmnopqrstuvwxyz"; //g++ main.cpp -std=c++17 -I . //cout < bool chmin(T& a, T b) { if (a > b) { a = b; return true; } else return false; } template bool chmax(T& a, T b) { if (a < b) { a = b; return true; } else return false; } // https://youtu.be/L8grWxBlIZ4?t=9858 // https://youtu.be/ERZuLAxZffQ?t=4807 : optimize // https://youtu.be/8uowVvQ_-Mo?t=1329 : division ll mod =998244353; //ll mod =1e9+7; struct mint { ll x; // typedef long long ll; mint(ll x=0):x((x%mod+mod)%mod){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod-a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;} mint operator+(const mint a) const { return mint(*this) += a;} mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= a;} mint pow(ll t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } // for prime mod mint inv() const { return pow(mod-2);} mint& operator/=(const mint a) { return *this *= a.inv();} mint operator/(const mint a) const { return mint(*this) /= a;} }; istream& operator>>(istream& is, const mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} // combination mod prime // https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619 struct combination { vector fact, ifact; combination(int n):fact(n+1),ifact(n+1) { //assert(n < mod); fact[0] = 1; for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i; ifact[n] = fact[n].inv(); for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i; } mint operator()(int n, int k) { if (k < 0 || k > n) return 0; if (n<0) return 0; return fact[n]*ifact[k]*ifact[n-k]; } mint p(int n, int k) { return fact[n]*ifact[n-k]; } } c(2000000); using vm=vector ; using vvm=vector ; struct edge{ int to; ll cost; edge(int to,ll cost) : to(to),cost(cost){} }; using ve=vector; using vve=vector; struct Compress{ vll a; map d,d2; int cnt; Compress() :cnt(0) {} void add(ll x){a.push_back(x);} void init(){ sets(a.begin(),a.end()); for(auto y:s)d[y]=cnt++; for(auto&y:a)y=d[y]; for(auto u:d)d2[u.second]=u.first; } ll to(ll x){return d[x];} //変換先 ll from(ll x){return d2[x];}//逆引き int size(){return cnt;} }; vll anss; const int N=105; mint dp[N][N*N][N]; mint dpsum[N][N*N][N]; void solve(int test){ int n,k; cin >> n >> k; vi a(n);rep(i,n)cin >> a[i]; Compress comp; rep(i,n)comp.add(a[i]); comp.init(); rep(i,n)a[i]=comp.to(a[i]); vi cnt(n); sort(a.rbegin(),a.rend()); dp[1][0][1]=1; dpsum[1][0][1]=1; { rep(j,n*(n-1)/2+1)repi(last,1,n+1)dpsum[1][j][last]+=dpsum[1][j][last-1]; } cnt[a[0]]++; repi(i,2,n+1){ rep(j,n*(n-1)/2+1)repi(last,1,i+1){ if(cnt[a[i-1]]==0){ int prej=j-(last-1); if(prej>=0)dp[i][j][last]+=dpsum[i-1][prej][n]; } else { int prej=j-(last-1-cnt[a[i-1]]); if(prej>=0 && last!=1)dp[i][j][last]+=dpsum[i-1][prej][last-1]; } } cnt[a[i-1]]++; rep(j,n*(n-1)/2+1)repi(last,1,n+1)dpsum[i][j][last]=dpsum[i][j][last-1]+dp[i][j][last]; } mint ans=dpsum[n][k][n]; cout << ans << endl; } //g++ main.cpp -std=c++17 -I . int main(){cin.tie(0);ios::sync_with_stdio(false); int t=1;//cin >> t; rep(test,t)solve(test); for(auto u:anss)cout << u << endl; }