#include <bits/stdc++.h>
#define mod 998244353ll
using namespace std;
using ll = long long;
int n, m;
ll qp(ll x, ll y) {
	ll ans = 1;
	while (y) {
		if (y & 1) ans = ans * x % mod;
		y >>= 1;
		x = x * x % mod;
	}
	return ans;
}
void ntt(vector<int>& a, int c, const vector<int>& rev, int lim) {
	static ll inv3 = qp(3, mod - 2);
	for (int i = 0; i < lim; ++i) if (i < rev[i]) swap(a[i], a[rev[i]]);
	for (int i = 1; i < lim; i <<= 1) {
		ll x = qp(~c ? 3 : inv3, (mod - 1) / (i << 1));
		for (int j = 0; j < lim; j += (i << 1)) {
			ll y = 1;
			for (int k = 0; k < i; ++k, y = y * x % mod) {
				ll p = a[j + k], q = a[j + k + i] * y % mod;
				a[j + k] = (p + q) % mod;
				a[j + k + i] = (p - q + mod) % mod;
			}
		}
	}
	if (c == -1) {
		ll inv = qp(lim, mod - 2);
		for (auto&& i : a) i = i * inv % mod;
	}
}
vector<int> mult(vector<int> a, vector<int> b) {
	int n = a.size(), m = b.size();
	int lim, l = 0;
	for (lim = 1; lim < n + m; lim <<= 1) ++l;
	a.resize(lim), b.resize(lim);
	vector<int> rev(lim);
	for (int i = 0; i < lim; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
	ntt(a, 1, rev, lim), ntt(b, 1, rev, lim);
	for (int i = 0; i < lim; ++i) a[i] = 1ll * a[i] * b[i] % mod;
	ntt(a, -1, rev, lim);
	a.resize(n + m - 1); //这一步很重要，在下面那道题中不加这个会 TLE！
	return a;
}
int main() {
	scanf("%d%d", &n, &m);
	vector<int> a(n + 1), b(m + 1);
	for (int i = 0; i <= n; ++i) {
		scanf("%d", &a[i]);
	}
	for (int j = 0; j <= m; ++j) {
		scanf("%d", &b[j]);
	}
	auto ans = mult(a, b);
	for (int i = 0; i <= (n + m); ++i) {
		printf("%d ", ans[i]);
	}
	return 0;
}