#include <bits/stdc++.h>
#ifdef LOCAL
#include <debug.hpp>
#else
#define debug(...) void(0)
#endif

#include "atcoder/convolution"
#include "atcoder/modint"

using namespace std;

typedef long long ll;
#define all(x) begin(x), end(x)
constexpr int INF = (1 << 30) - 1;
constexpr long long IINF = (1LL << 60) - 1;
constexpr int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

template <class T> istream& operator>>(istream& is, vector<T>& v) {
    for (auto& x : v) is >> x;
    return is;
}

template <class T> ostream& operator<<(ostream& os, const vector<T>& v) {
    auto sep = "";
    for (const auto& x : v) os << exchange(sep, " ") << x;
    return os;
}

template <class T, class U = T> bool chmin(T& x, U&& y) { return y < x and (x = forward<U>(y), true); }

template <class T, class U = T> bool chmax(T& x, U&& y) { return x < y and (x = forward<U>(y), true); }

template <class T> void mkuni(vector<T>& v) {
    sort(begin(v), end(v));
    v.erase(unique(begin(v), end(v)), end(v));
}

template <class T> int lwb(const vector<T>& v, const T& x) { return lower_bound(begin(v), end(v), x) - begin(v); }

/*
(j + k)^3 = j^3 + 3(j + k)jk + k^3
(j + k)jk = ((j + k)^3 - j^3 - k^3) / 3
B_j C_k x^{(j + k)jk} = (B_j x^{-j^3}) (C_k x^{-k^3}) x^{(j + k)^3}
(j + k)^3 // 3 - j^3 // 3 - k^3 // 3 + ((j + k) % 3 - j % 3 - k % 3) / 3
最後の項が -1 or 0
*/

using mint = atcoder::static_modint<943718401>;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, x;
    cin >> n >> x;
    vector<int> A(n + 1), B(n + 1), C(n + 1);
    cin >> A >> B >> C;

    mint ans = 0;
    if (x == 0) {
        for (int i = 0; i <= n; i++) {
            ans += A[i] * B[0] * C[i];
            ans += A[i] * B[i] * C[0];
        }
        ans -= A[0] * B[0] * C[0];
        cout << ans.val() << '\n';
        return 0;
    }

    vector<mint> power(n + 1);
    for (int i = 0; i <= n; i++) power[i] = mint(x).pow(1LL * i * i * i / 3);
    mint inv = mint(x).inv();
    vector<mint> b(n + 1, 0), c(n + 1, 0);
    for (int j = 0; j < 3; j++) {
        for (int k = 0; k < 3; k++) {
            for (int i = 0; i <= n; i++) {
                mint p = power[i].inv();
                b[i] = (i % 3 == j ? p * B[i] : 0);
                c[i] = (i % 3 == k ? p * C[i] : 0);
            }
            auto a = atcoder::convolution(b, c);
            for (int i = 0; i <= n; i++) ans += a[i] * power[i] * A[i] * (j + k >= 3 ? inv : 1);
        }
    }

    cout << ans.val() << '\n';
    return 0;
}