#include using namespace std; using ll = long long; #define rep(i,n) for (int i = 0; i < (n); ++i) #define rep2(i,m,n) for (int i = m; i < (n); ++i) #define all(x) (x).begin(),(x).end() inline int popcount(const int x) { return __builtin_popcount(x);} template void chmin(T &a, const T &b) noexcept { if (b < a) a = b;} template void chmax(T &a, const T &b) noexcept { if (a < b) a = b;} template void drop(const T &x) { std::cout< void debug_out(const T &x, const Args &... args) { std::cout<= 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;} bool operator==(const mint rhs) const { return x == rhs.x; } bool operator!=(const mint rhs) const { return x != rhs.x; } bool operator<(const mint &a) const{ return x>(istream& is, mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} template struct Combination { map, mint> ncr_mp; Combination(){} ModInt C(ll n, ll r) { if (r < 0 || n < r) return ModInt(0); auto key = pair(n,r); if (ncr_mp.count(key)) return ncr_mp[key]; if (r == 0 or r == n) return ModInt(1); if (r == 1 or r == n-1) return ModInt(n); //return mp[key] = nCr(n,r-1) * (n-r+1) / r; return ncr_mp[key] = C(n-1,r) * n / (n-r); } ModInt H(ll n, ll r) { if (r < 0 || n < 0) return 0; return r == 0 ? ModInt(1) : C(n+r-1, r); } ModInt operator()(ll n, ll k) { return C(n, k);} }; int main() { ll n, k; cin >> n >> k; vector a(n); rep(i,n) cin >> a[i]; mint ans = 0; Combination comb; rep(i, n) { //i番目が残るようなlの選び方はi+1個からk個,rの選び方はn-i個からk個取るパターン数と同じ. ans += comb.H(i+1, k) * comb.H(n-(i+1)+1, k) * mint(a[i]); } cout << ans << endl; return 0; }