//#define //_GLIBCXX_DEBUG #include #include #include using namespace std; using namespace atcoder; using ll = long long; using ull = unsigned long long; using vll=vector; using vvll=vector>; using Graph=vvll; using Edgegraph=vector>>; using vch=vector; using vvch=vector>; using P=pair; using vP=vector

; using tup=tuple; using vbl=vector; using vvbl=vector; using vs=vector; using vvs=vector; using vd=vector; using vvd=vector; using mint = atcoder::modint998244353; const int infint = 1073741823; const ll inf = 1LL << 60; template inline bool chmax(T& a,T b){if (a inline bool chmin(T& a,T b){if (a>b){a=b;return 1;}return 0;} #define rep(i,x,lim) for(ll i = (x);i < (ll)(lim);i++) #define rep2(j,x,lim) for(int j = (x);j < (int)(lim);j++) const ll big=(1e+9)+7; const ll big2=998244353; ll dx[8]={1,-1,0,0,1,1,-1,-1}; ll dy[8]={0,0,1,-1,1,-1,1,-1}; int modpow(ll x,ll n,ll m){ if(n==0) return 1%m; x=((x%m)+m)%m; if(n%2==0){ ll r=modpow(x,n/2,m); return r*r%m; } else{ ll r=modpow(x,n/2,m); return r*r%m*x%m; } } //pは素数でなければならない。 int revmod(ll x,ll p){return modpow(x,p-2,p);} //99 のRを高速に求める int modp(ll p,ll q){ ll gc=gcd(p,q); p/=gc;q/=gc; ll rev=revmod(p,big2); return (rev*q)%big2; } //nCrを求める modbig2 int nCr(ll n,ll r){ ll ans=1; rep(i,1,n+1) ans=(ans*i)%big2; rep(i,1,r+1) ans=(ans*modpow(i,big2-2,big2))%big2; rep(i,1,n-r+1) ans=(ans*modpow(i,big2-2,big2))%big2; return ans; } ll op(ll a,ll b){return max(a,b);} ll opmin(ll a,ll b){return min(a,b);} ll e(){return 0;} template size_t HashCombine(const size_t seed,const T &v){ return seed^(std::hash()(v)+0x9e3779b9+(seed<<6)+(seed>>2)); } /* pair用 */ template struct std::hash>{ size_t operator()(const std::pair &keyval) const noexcept { return HashCombine(std::hash()(keyval.first), keyval.second); } }; const int MAX = 510000; mint fac[MAX], finv[MAX], inv[MAX]; void COMinit() { const int MOD = mint::mod(); fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; for (int i = 2; i < MAX; i++){ fac[i] = fac[i - 1] * i; inv[i] = MOD - inv[MOD%i] * (MOD / i); finv[i] = finv[i - 1] * inv[i]; } } // 二項係数計算 mint COM(int n, int k){ if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * finv[k] * finv[n - k]; } int main(){ ll N; cin >> N; ll M,T; cin >> M >> T; vll A(M); vll B(N); rep(i,0,M){ cin >> A[i]; B[A[i]-1]++; } ll up=1e+15; ll down=-1; while(up-down>1){ ll mid=(up+down)/2; vll C(N); ll amari=M; rep(i,0,N){ C[i]+=min(B[i],mid); amari-=min(B[i],mid); } rep(i,0,N){ amari-=(mid-C[i])/T; } if(amari<=0) up=mid; else down=mid; } cout << up; }