// Skyqwq
#include <iostream>
#include <cstdio>
#include <cstring>

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

using namespace std;

typedef long long LL;

template <typename T> void chkMax(T &x, T y) { if (y > x) x = y; }
template <typename T> void chkMin(T &x, T y) { if (y < x) x = y; }

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;
}
 
template <typename T> void print(T x) {
	if (x < 0) { putchar('-'); print(-x); return ; }
    if (x >= 10) print(x / 10);
    putchar((x % 10) + '0');
}

const int N = 1005, P = 998244353;

int n, K, x, y, p, pw[N], f[N][N], s[N][N], ans;

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

int inline power(int a, int b) {
	int res = 1;
	while (b) {
		if (b & 1) res = (LL)res * a % P;
		a = (LL)a * a % P;
		b >>= 1;
	}
	return res;
}

void inline clear() {
	memset(pw, 0, sizeof pw);
	memset(f, 0, sizeof f);
	memset(s, 0, sizeof s);
}

void work1() {
	for (int i = 0; i <= K; i++)
		pw[i] = (LL)power(p, i) * (1 + P - p) % P;
	for (int i = 0; i <= K + 1; i++) s[0][i] = 1;
	int l = min(n, K);
	for (int i = 1; i <= l; i++) {
		int lim = i == 0 ? K : K / i;
		for (int j = 0; j <= lim; j++) {
			int v = 0;
			for (int x = 1; x <= i; x++)
				add(v, (LL)s[x - 1][j + 1] * s[i - x][j] % P);
			f[i][j] = (LL)pw[j] * v % P;
		}
		for (int j = lim; ~j; j--)
			s[i][j] = (s[i][j + 1] + f[i][j]) % P;
	}
}

int F[N << 1], G[N << 1], H[N << 1], C[N << 1];

void inline mul(int *f, int *g, int *h) {
	int l = 2 * K + 2;
	for (int i = 0; i <= l; i++) h[i] = 0;
	for (int i = 0; i <= K + 1; i++) {
		for (int j = 0; j <= K + 1; j++) {
			add(h[i + j], (LL)f[i] * g[j] % P);
		}
	}
}

int inline work2() {
	memset(F, 0, sizeof F);
	memset(G, 0, sizeof G);
	for (int i = 0; i <= K; i++) F[i] = s[i][0];
	for (int i = 1; i <= K + 1; i++) G[i] = (LL)(1 - p + P) * s[i - 1][1] % P;
	for (int i = 1; i <= K + 1; i++) G[i] = (P - G[i]) % P;
	add(G[0], 1);
	mul(F, G, H);
	for (int i = K + 1; i <= 2 * K + 2; i++) H[i] = 0;
	int b = n;
	while (b) {
		for (int i = 0; i <= K + 1; i++) 
			F[i] = (i & 1) ? (P - G[i]) % P : G[i];
		mul(F, G, C);
		for (int i = 0; i <= K + 1; i++)
			G[i] = C[i * 2];
		mul(F, H, C);
		for (int i = 0; i <= K; i++)
			H[i] = C[i * 2 + (b & 1)];
		b >>= 1;
	}
	return H[0];
}

int inline solve() {
	if (K == -1) return 0;
	if (K == 0) return power((1 - p + P) % P, n);
	clear();
	work1();
	if (n <= K) return s[n][0];
	else return work2();
}
 
int main() {
	read(n), read(K), read(x), read(y);
	p = (LL)x * power(y, P - 2) % P;
	int ans = solve();
	K--;
	ans = (ans - solve() + P) % P;
	printf("%d\n", ans);
	return 0;
}