#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for (int i=(int)(a);i<(int)(b);i++)

#include <atcoder/modint>
using mint = atcoder::modint998244353;

struct F{
    mint c = 0;
    mint v = 0;
};

F operator*=(F &a, F b){
    return a = {a.c * b.c, a.c * b.v + a.v * b.c};
}
F operator+=(F &a, F b){
    return a = {a.c + b.c, b.v + a.v};
}

void solve();
// DEAR MYSTERIES / TOMOO
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1;
    // cin >> t;
    rep(i, 0, t) solve();
}

void solve(){
    int N, K;
    cin >> N >> K;
    vector<int> A(N);
    rep(i, 0, N) cin >> A[i];
    vector<F> dp1(K + 1), dp2(K + 1);
    dp1[0] = {1, 0};
    rep(rp, 0, N){
        for (int i = K; i > 0; i--){
            dp2[i - 1] += dp2[i];
        }
        dp2[K] += dp1[K];
        dp1[K] = {0, 0};
        rep(i, 0, K){
            dp1[i + 1] += dp1[i];
        }
        mint p2 = 1, pa = 1;
        rep(i, 0, K + 1){
            F tmp = {p2, pa * p2};
            dp1[i] *= tmp;
            dp2[i] *= tmp;
            p2 *= 2, pa *= A[rp];
        }
    }
    mint ans = 0;
    rep(i, 0, K + 1) ans += dp2[i].v;
    ans /= (mint(4)).pow(K);
    cout << ans.val() << "\n";
}