#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const int mod = 258280327;

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma") 
namespace {
    template<int n, typename T>
    void mult(const T *__restrict a, const T *__restrict b, T *__restrict res) {
        if (n <= 64) { // if length is small then naive multiplication if faster
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                    res[i + j] %= mod;
                }
            }
        } else {
            // cout << n << endl;
            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]; // atmp(x) - sum of two halfs a(x)
                atmp[i] %= mod;
                btmp[i] = b[i] + b[i + mid]; // btmp(x) - sum of two halfs b(x)
                btmp[i] %= mod;
            }
            // cout << "sum" << endl;
            mult<mid>(atmp, btmp, E); // Calculate E(x) = (alow(x) + ahigh(x)) * (blow(x) + bhigh(x))
            // cout << "mult1" << endl;
            mult<mid>(a + 0, b + 0, res); // Calculate rlow(x) = alow(x) * blow(x)
            // cout << "mult2" << endl;
            mult<mid>(a + mid, b + mid, res + n); // Calculate rhigh(x) = ahigh(x) * bhigh(x)
            // cout << "mult3" << endl;
            for (int i = 0; i < mid; i++) { // Then, calculate rmid(x) = E(x) - rlow(x) - rhigh(x) and write in memory
                const auto tmp = res[i + mid];
                res[i + mid] += E[i] - res[i] - res[i + 2 * mid];
                res[i + mid] %= mod;
                res[i + 2 * mid] += E[i + mid] - tmp - res[i + 3 * mid];
                res[i + 2 * mid] %= mod;
            }
            // cout << "done" << endl;
        }
    }
}

const int nmax = (1 << 18);
ll a[nmax],b[nmax],ret[nmax];

int main(){
    int n,m;
    cin >> n;
    for(int i = 0;i <= n;++i) cin >> a[i];
    cin >> m;
    for(int i = 0;i <= m;++i) cin >> b[i];
    int sz = 1;
    while((1 << sz) < max(n,m)) ++sz;
    mult<nmax, ll>(a, b, ret);
    cout << n + m << endl;
    for(int i = 0;i <= n + m;++i){
        auto x = ret[i];
        if(ret[i] < 0) ret[i] += mod;
        cout << x << ' ';
    }
    cout << endl;
    return 0;
}