#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef complex<ld> compd;
#define quickIO()	{cin.tie(0);	cout.sync_with_stdio(false);}
#define reach(i,a)	for(auto i:a)
#define rep(i,n)	for(int i=0;i<((int)n);i++)
#define REP(i,n)	for(int i=0;i<=((int)n);i++)
#define srep(i,a,n)	for(int i=a;i<((int)n);i++)
#define SREP(i,a,n)	for(int i=a;i<=((int)n);i++)
#define rrep(i,n)	for(int i=n-1;i>=0;i--)
#define RREP(i,n)	for(int i=n;i>=0;i--)
#define all(a)	(a).begin(),(a).end()
#define mp(a,b)	make_pair(a,b)
#define mt	make_tuple
#define pb	push_back
template<typename T> istream& operator >> (istream& is, vector<T>& vec) {
	for (T& x : vec)	is >> x;
	return is;
}
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
	os << "[";
	rep(i, vec.size())	os << (i ? ", " : "") << vec[i];
	os << "]";
	return os;
}
template<typename T> istream& operator >> (istream& is, pair<T, T>& p) {
	is >> p.first >> p.second;
	return is;
}
template<typename T> ostream& operator << (ostream& os, pair<T, T>& p) {
	os << "(" << p.first << ", " << p.second << ")";
	return os;
}
int bitcnt(ll x) {
	x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) + (x & 0x5555555555555555);
	x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) + (x & 0x3333333333333333);
	x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) + (x & 0x0F0F0F0F0F0F0F0F);
	x = ((x & 0xFF00FF00FF00FF00) >> 8) + (x & 0x00FF00FF00FF00FF);
	x = ((x & 0xFFFF0000FFFF0000) >> 16) + (x & 0x0000FFFF0000FFFF);
	x = ((x & 0xFFFFFFFF00000000) >> 32) + (x & 0x00000000FFFFFFFF);
	return x;
}
int bitcnt(int x) {
	x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555);
	x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333);
	x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F);
	x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF);
	x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF);
	return x;
}
ll sqrtll(ll x) {
	ll left = 0, right = x;
	rep(i, 100) {
		ll mid = (left + right) >> 1;
		if (mid*mid <= x)	left = mid;
		else	right = mid;
	}
	return left;
}
ll gcd(ll a, ll b) {
	return a % b == 0 ? b : gcd(b, a%b);
}
#define debug(x)	printf("Case #%d: ", x)
#define DEBUG 0
const ll inf = 1e18;
const ld infl = 1e100;
const ll mod = 998244353;
const ld eps = 1e-9;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };

ll fact[300010];
ll invfact[300010];

ll modpow(ll a, ll b) {
	ll ret = 1;
	while (b) {
		if (b & 1)	ret = (ret*a) % mod;
		a = (a*a) % mod;
		b /= 2;
	}
	return ret;
}
ll nCm(int n, int m) {
	return (fact[n] * (invfact[m] * invfact[n - m] % mod) % mod);
}

int main() {
	// 赤と青を別々に塗って、たまたま重なったところを緑としておけば独立に計算してOK
	// 赤のブロック数をiとすると、青のブロックで稼ぐ必要があるのはK-A*i点。Bで割り切った商がjとすると、
	// NCi * NCj通りが該当する。
	ll n, a, b, k;	cin >> n >> a >> b >> k;
	fact[0] = invfact[0] = 1;
	SREP(i, 1, n) {
		fact[i] = (fact[i - 1] * i) % mod;
		invfact[i] = (invfact[i - 1] * modpow(i, mod - 2)) % mod;
	}
	ll ret = 0;
	REP(i, n) {
		ll left = k - a * i;
		if (left%b != 0)	continue;
		ll j = left / b;
		if (j < 0 || n < j)	continue;
		(ret += nCm(n, i) * nCm(n, j) % mod) %= mod;
	}
	cout << ret << endl;
	return 0;
}

Measured memory_bytes: 4992