/** * @FileName a.cpp * @Author kanpurin * @Created 2021.06.20 23:10:04 **/ #include "bits/stdc++.h" using namespace std; typedef long long ll; template< int MOD > struct mint { public: long long x; mint(long long x = 0) :x((x%MOD+MOD)%MOD) {} mint(std::string &s) { long long z = 0; for (int i = 0; i < s.size(); i++) { z *= 10; z += s[i] - '0'; z %= MOD; } this->x = z; } 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) { long long n = MOD - 2; mint u = 1, b = a; while (n > 0) { if (n & 1) { u *= b; } b *= b; n >>= 1; } return *this *= u; } mint operator+(const mint &a) const { mint res(*this); return res += a; } mint operator-() const {return mint() -= *this; } mint operator-(const mint &a) const { mint res(*this); return res -= a; } mint operator*(const mint &a) const { mint res(*this); return res *= a; } mint operator/(const mint &a) const { mint res(*this); return res /= a; } friend std::ostream& operator<<(std::ostream &os, const mint &n) { return os << n.x; } friend std::istream &operator>>(std::istream &is, mint &n) { long long x; is >> x; n = mint(x); return is; } bool operator==(const mint &a) const { return this->x == a.x; } bool operator!=(const mint &a) const { return this->x != a.x; } mint pow(long long k) const { mint ret = 1; mint p = this->x; while (k > 0) { if (k & 1) { ret *= p; } p *= p; k >>= 1; } return ret; } }; constexpr int MOD = 998244353; int main() { int t;cin >> t; using T = pair; while(t--) { int n,k;cin >> n >> k; vector a(n); mint ans = 1; for (int i = 2; i <= k; i++) ans *= i; int sum = 0; for (int i = 0; i < n; i++) { cin >> a[i]; sum += a[i]; } priority_queue,function> pq([&](T x,T y){return a[x.first]*y.second < a[y.first]*x.second;}); for (int i = 0; i < n; i++) pq.push({i,1}); while(k--) { auto p = pq.top(); pq.pop(); pq.push({p.first,p.second+1}); ans *= mint(a[p.first])/sum/p.second; } cout << ans << endl; } return 0; }