#pragma GCC optimization ("O3") #include using namespace std; using ll = long long; using vec = vector; using mat = vector; using pll = pair; #define INF (1LL<<61) #define MOD 1000000007LL //#define MOD 998244353LL #define EPS (1e-10) #define PR(x) cout << (x) << endl #define PS(x) cout << (x) << " " #define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i)) #define FORE(i,v) for(auto (i):v) #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((ll)(x).size()) #define REV(x) reverse(ALL((x))) #define ASC(x) sort(ALL((x))) #define DESC(x) {ASC((x)); REV((x));} #define BIT(s,i) (((s)>>(i))&1) #define pb push_back #define fi first #define se second template inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;} template inline int chmax(T& a, T b) {if(a=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 {mint b(*this); return b+=a;} mint operator-(const mint& a) const {mint b(*this); return b-=a;} mint operator*(const mint& a) const {mint b(*this); return b*=a;} mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;} mint inv() const {return pow(MOD-2);} mint& operator/=(const mint& a) {return *this*=a.inv();} mint operator/(const mint& a) const {mint b(*this); return b/=a;} }; istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;} ostream &operator<<(ostream& os, const mint& a) {return os<; using mmat = vector; using dvec = vector; using dmat = vector; int main() { ll N, M; mint P; cin >> N >> M >> P; P /= mint(100); vec V(N); REP(i,0,N) cin >> V[i]; DESC(V); mvec S(N+1), F(N+M+1); REP(i,1,N+1) S[i] = S[i-1]+mint(V[i-1]); F[0] = mint(1); REP(i,1,N+M+1) F[i] = mint(i)*F[i-1]; auto binom = [&](ll n, ll k) -> mint { if(n < 0 || k < 0 || n < k) return mint(0); return F[n]*F[k].inv()*F[n-k].inv(); }; mint ans, Q = mint(1)-P, R; REP(i,1,N+1) { mint t = binom(M+i-2, i-1)*P.pow(M)*Q.pow(i-1); ans += t*S[i-1]; R += t; } ans += (mint(1)-R)*S[N]; PR(ans.x); return 0; } /* */