#define GCC Optimaze("Ofast")
#include <bits/stdc++.h>
#define all(x) (x).begin(),(x).end()

using namespace std;
using ll = long long;

const int LOG = 18;
const int N = 100100;
const ll mod = 65537;

int normalize(ll x) {
    x %= mod;
    x += mod;
    x %= mod;
    return x;
}

int add(ll a, ll b) {
    return normalize(a + b);
}
int sub(ll a, ll b) {
    return normalize(a - b);
}

ll mult(ll a, ll b) {
    a = normalize(a);
    b = normalize(b);
    return normalize(a * b);
}

ll fastPow(ll a, ll p) {
    ll ans = 1;
    a = normalize(a);
    while (p) {
        if (p & 1) {
            ans *= a;
            ans %= mod;
        }
        a *= a;
        a %= mod;
        p /= 2;
    }
    return normalize(ans);
}

ll inv(ll a) {
    return fastPow(a, mod - 2);
}


namespace {
template<int n, typename T = int>
void mult(const T *__restrict a, const T *__restrict b, T *__restrict res) {
    if (n <= 64) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                res[i + j] += a[i] * 1ll * b[j] % mod;
                res[i + j] = normalize(res[i + j]);
            }
        }
    } else {
        const int mid = n / 2;
        alignas(64) T btmp[n], E[n] = {};
        auto atmp = btmp + mid;
        for (int i = 0; i < mid; i++) {
            atmp[i] = (a[i] + a[i + mid]) % mod;
            btmp[i] = (b[i] + b[i + mid]) % mod;
        }
        mult<mid>(atmp, btmp, E);
        mult<mid>(a + 0, b + 0, res);
        mult<mid>(a + mid, b + mid, res + n);
        for (int i = 0; i < mid; i++) {
            const auto tmp = res[i + mid];
            res[i + mid] += E[i] - res[i] - res[i + 2 * mid];
            res[i + 2 * mid] += E[i + mid] - tmp - res[i + 3 * mid];
            res[i + mid] = normalize(res[i + mid]);
            res[i + 2 * mid] = normalize(res[i + 2 * mid]);
        }
    }
}
}

static alignas(64) int A[N], B[N], ans[N], fact[N], invfact[N];


void Solve() {
    int n, k;
    cin >> n >> k;
    k = normalize(k);
    vector<ll> b(n);
    memset(A, 0, sizeof A);
    memset(B, 0, sizeof B);
    memset(ans, 0, sizeof ans);
    for (int i = 0; i < n; i++) cin >> b[i];
    for (int i = 0; i < n; i++) {
        b[i] = normalize(b[i]);
    }
    for (int i = 0; i < n; i++) {
        B[i] = mult(b[n - 1 - i], invfact[i]);
        A[i] = mult(fastPow(normalize(-k), i), invfact[i]);
    }
    const int NMAX = 1 << LOG;
    mult<NMAX>(A, B, ans);
    for (int i = 0; i < n; i++) {
        cout << mult(ans[i], fact[i]) << " ";
    }
    cout << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    fact[0] = 1;
    invfact[0] = 1;
    for (int i = 1; i < N; i++) {
        fact[i] = mult(fact[i - 1], i);
        invfact[i] = inv(fact[i]);
    }
    int t;
    cin >> t;
    while (t--) {
        Solve();
    }
}