// Skyqwq
#include <bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define mp make_pair

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }

template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

const int X = 21, Y = 13, P = 998244353, O = 998244352;

typedef vector<int> Poly;

#define pb push_back

const int N = 5000000 + 5, G = 3;

int A[N], rev[N], mod;
int lim = 1, len = 0, W[24][N];

int inline power(int a, int b, int Mod = P) {
	int res = 1;
	while (b) {
		if (b & 1) res = (LL)res * a % Mod;
		a = (LL)a * a % Mod;
		b >>= 1;
	}
	return res;
}
int Gi = power(G, P - 2, P), inv2 = power(2, P - 2, P);


void inline NTT(int c[], int lim, int o) {
	for (int i = 0; i < lim; i++)
		if (i < rev[i]) swap(c[i], c[rev[i]]);
	for (int k = 1, t = 0; k < lim; k <<= 1, t++) {
		for (int i = 0; i < lim; i += (k << 1)) {
			for (int j = 0; j < k; j++) {
				int u = c[i + j], v = (LL)c[i + k + j] * W[t][j] % P;
				c[i + j] = u + v >= P ? u + v - P : u + v;
				c[i + j + k] = u - v < 0 ? u - v + P : u - v; 
			}
		}
	}
	if (o == -1) {
		reverse(c + 1, c + lim);
		int inv = power(lim, P - 2, P);
		for (int i = 0; i < lim; i++)
			c[i] = (LL)c[i] * inv % P;
	}
}

void inline setN(int n) {
	lim = 1, len = 0;
	while (lim < n) lim <<= 1, len++;
	for (int i = 0; i < lim; i++)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
}

Poly inline NTT(Poly a, int o) {
	int n = a.size();
	for (int i = 0; i < n; i++) A[i] = a[i];
	NTT(A, lim, o);
	a.clear();
	for (int i = 0; i < lim; i++) a.push_back(A[i]), A[i] = 0;
	return a;
}


Poly inline mul (Poly a, Poly b, int newn = -1) {
	if (newn == -1) newn = a.size() + b.size() - 1;
	setN(a.size() + b.size() - 1);
	Poly c = NTT(a, 1), d = NTT(b, 1);
	for (int i = 0; i < lim; i++) c[i] = (LL)c[i] * d[i] % P;
	d = NTT(c, -1); d.resize(newn);
	return d;
}
// 用到的最大的 n
void inline init(int n) {
	setN(2 * n);
	for (int k = 1, t = 0; k < lim; k <<= 1, t++) {
		int wn = power(G, (P - 1) / (k << 1));
		W[t][0] = 1;
		for (int j = 1; j < k; j++) W[t][j] = (LL)W[t][j - 1] * wn % P;
	}
}

int x, y, z, U, V;

int cnt[20000000];

int v1[20000000];

LL n;

int ans;

void inline add(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}

int tp(LL x) {
	int c = 0;
	while (x) c += x % 3, x /= 3;
	return c;
}

void inline del(int &x, int y) {
	x -= y;
	if (x < 0) x += P;
}

void inline work(LL l, LL r) {
	for (LL i = l; i <= r; i++) {
		add(ans, 1ll * power(x, i % O) * power(y, __builtin_popcountll(i)) % P * power(z, tp(i)) % P);
	}
}

int main() {
    U = 1 << X;
    V = 1;
    ans = P - 1;
    for (int i = 0; i < Y; i++) V *= 3;
    init(V);
    read(n), read(x), read(y), read(z);
	Poly a(U + 1, 0), b(V, 0);
	int val = 1;
	cnt[0] = 1;
	for (int i = 0; i < U; i++) {
		cnt[i] = 1ll * cnt[i >> 1] * ((i & 1) ? y : 1) % P;
		a[U - i] = 1ll * cnt[i] * val % P;
		val = 1ll * val * x % P;
	}
	cnt[0] = 1;
	int z2 = 1ll * z * z % P;
	for (int i = 0; i < V; i++) {
		cnt[i] = cnt[i / 3];
		if (i % 3 == 2) cnt[i] = 1ll * cnt[i] * z2 % P;
		else if (i % 3 == 1) cnt[i] = 1ll * cnt[i] * z % P;
		b[i] = cnt[i];
	}
	Poly c = mul(a, b);

	int kx = power(x, U);

	int p1 = 0;
	int vc = 1;

	cnt[0] = v1[0] = 1;

	for (LL l = 0; l <= n; ) {
		int m1 = l / U, m2 = l / V;
		cnt[m1] = 1ll * cnt[m1 >> 1] * ((m1 & 1) ? y : 1) % P;
		v1[m2] = v1[m2 / 3];
		if (m2 % 3 == 2) v1[m2] = 1ll * v1[m2] * z2 % P;
		else if (m2 % 3 == 1) v1[m2] = 1ll * v1[m2] * z % P;
		if (p1 < m1) ++p1, vc = 1ll * vc * kx % P;
		assert(p1 == m1);
		LL n1 = (m1 + 1ll) * U - 1, n2 = (m2 + 1ll) * V - 1;
		LL nx = min(n1, n2);
		if (nx > n) {
			work(l, n);
			break;
		}
		
		LL dt = m1 * U - m2 * V + U;
		if (dt >= 0 && dt < c.size())
			add(ans, 1ll * c[dt] * vc % P * cnt[m1] % P * v1[m2] % P);
		else assert(false);		
		l = nx + 1;
		
	}
	printf("%d\n", ans);
    return 0;
}