Problem:
Score : 700 points

Problem Statement

Takahashi has a tower which is divided into N layers.
Initially, all the layers are uncolored. Takahashi is going to paint some of the layers in red, green or blue to make a beautiful tower.
He defines the beauty of the tower as follows:

The beauty of the tower is the sum of the scores of the N layers, where the score of a layer is A if the layer is painted red, A+B if the layer is painted green, B if the layer is painted blue, and 0 if the layer is uncolored.

Here, A and B are positive integer constants given beforehand. Also note that a layer may not be painted in two or more colors.

Takahashi is planning to paint the tower so that the beauty of the tower becomes exactly K.
How many such ways are there to paint the tower? Find the count modulo 998244353.
Two ways to paint the tower are considered different when there exists a layer that is painted in different colors, or a layer that is painted in some color in one of the ways and not in the other.

Constraints

1 ≤ N ≤ 3×10^5

1 ≤ A,B ≤ 3×10^5

0 ≤ K ≤ 18×10^{10}

All values in the input are integers.

Input

Input is given from Standard Input in the following format:

N A B K

Output

Print the number of the ways to paint tiles, modulo 998244353.

Sample Input 1

4 1 2 5

Sample Output 1

40

In this case, a red layer worth 1 points, a green layer worth 3 points and the blue layer worth 2 points. The beauty of the tower is 5 when we have one of the following sets of painted layers:

1 green, 1 blue

1 red, 2 blues

2 reds, 1 green

3 reds, 1 blue

The total number of the ways to produce them is 40.

Sample Input 2

2 5 6 0

Sample Output 2

1

The beauty of the tower is 0 only when all the layers are uncolored. Thus, the answer is 1.

Sample Input 3

90081 33447 90629 6391049189

Sample Output 3

577742975

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma region Template
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
#include <utility>
#include <stack>
#include <set>
#include <algorithm>
#include <bitset>
#include <functional>
#include <ctime>
#include <cassert>
#include <valarray>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <complex>
#include <regex>
#include <numeric>
#include <typeinfo>
#include <type_traits>
#include <fstream>
#pragma comment(linker, "/STACK:167772160")

using namespace std;

#ifdef LOCAL

template <typename T, typename Q>
ostream& operator<<(ostream& os, pair<T, Q> p)
{
	return (os << '(' << p.first << ", " << p.second << ')');
}

int _iterable_indent = 0;
template<class T>
std::ostream& print_iterable(std::ostream& os, T beg, T end)
{
	if (_iterable_indent > 0)
		os << "\n";
	for (int i = 0; i < _iterable_indent; ++i)
		os << "\t";
	os << "[";
	for (auto it = beg; it != end; ++it)
	{
		if (it != beg)
			os << ", ";
		++_iterable_indent;
		os << *it;
		--_iterable_indent;
	}
	os << "]";
	return os;
}

template <class T >
std::ostream& operator << (std::ostream& os, const std::vector<T>& v)
{
	return print_iterable(os, v.begin(), v.end());
}
template <class T >
std::ostream& operator << (std::ostream& os, const std::set<T>& v)
{
	return print_iterable(os, v.begin(), v.end());
}
template <class T >
std::ostream& operator << (std::ostream& os, const std::multiset<T>& v)
{
	return print_iterable(os, v.begin(), v.end());
}
template <class T >
std::ostream& operator << (std::ostream& os, const std::unordered_set<T>& v)
{
	return print_iterable(os, v.begin(), v.end());
}
template <class T, class Q>
std::ostream& operator << (std::ostream& os, const std::map<T, Q>& v)
{
	return print_iterable(os, v.begin(), v.end());
}
template <class T, class Q>
std::ostream& operator << (std::ostream& os, const std::unordered_map<T, Q>& v)
{
	return print_iterable(os, v.begin(), v.end());
}

vector<string> _split(const string& s, char c) {
	vector<string> v;
	stringstream ss(s);
	string x;
	auto arr_skip = false;
	while (getline(ss, x, c))
	{
		while (x[0] == ' ')
			x = x.substr(1);
		if (arr_skip)
		{
			for (auto h : x)
				if (h == ')')
					arr_skip = false;
			continue;
		}
		if (x.size() >= 3 && x.substr(0, 3) == "ARR")
		{
			arr_skip = true;
			v.emplace_back(x.substr(4));
		}
		else
			v.emplace_back(x);
	}
	return v;
}

void _print(vector<string>::iterator) {}
template<typename T, typename... Args>
void _print(vector<string>::iterator it, T a, Args... args) {
	string name = it->substr((*it)[0] == ' ', it->length());
	if (isalpha(name[0]))
		cerr << name << " = " << a << endl;
	else
		cerr << name << endl;
	_print(++it, args...);
}

template<class T>
auto _kekeke(T x)
{
	return x;
}

template<class T>
auto make_vector(T arr)
{
	return arr;
}

template<class T, class... Args>
auto make_vector(T arr, int len, Args... args)
{
	auto inside = make_vector(arr[0], args...);
	vector<decltype(_kekeke(inside))> v;
	for (int i = 0; i < len; ++i)
		v.push_back(make_vector(arr[i], args...));
	return v;
}

#define dbg(...) { _print(_split(#__VA_ARGS__, ',').begin(), __VA_ARGS__); }
#define ARR(arr, ...) make_vector(arr, __VA_ARGS__)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define dbg(...) (void)0
#define ARR(arr, ...) (void)0
#define eprintf(...) (void)0
#endif
#define mp make_pair
#define rand govno_ebanoe
#define dbg_time() eprintf("\n\nTime: %.3lf\n", double(clock()) / CLOCKS_PER_SEC)

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;

const int INF = (int)1e9 + 10;
const ll LINF = ll(2e18) + 10;
const double PI = acosl(-1);
const double eps = 1e-8;
#pragma endregion

const int P = 998244353;
const int N = 3e5 + 10;
int fact[N];
int rfact[N];
int mul(int a, int b)
{
	return (ll)a * b % P;
}
int bpow(int x, int p)
{
	int res = 1;
	for (; p > 0; p /= 2, x = mul(x, x))
		if (p % 2 == 1)
			res = mul(res, x);
	return res;
}

int cnk(int n, int k)
{
	if (k < 0 || k > n)
		return 0;
	return mul(fact[n], mul(rfact[k], rfact[n - k]));
}

ll getc(ll n, ll x, ll y)
{
	return mul(cnk(n, x), cnk(n, y));
}

void solve()
{
	fact[0] = 1;
	for (int i = 1; i < N; ++i)
		fact[i] = mul(fact[i - 1], i);
	rfact[N - 1] = bpow(fact[N - 1], P - 2);
	for (int i = N - 2; i >= 0; --i)
		rfact[i] = mul(rfact[i + 1], i + 1);
	int n, a, b;
	ll k;
	cin >> n >> a >> b >> k;
	ll ans = 0;
	for (ll x = 0; x <= n; ++x)
	{
		auto left = k - a * x;
		if (left < 0 || left % b != 0)
			continue;
		auto y = left / b;
		ans += getc(n, x, y);
		ans %= P;
	}
	cout << ans << endl;
}

#ifdef LOCAL
#define ERR_CATCH
//#define NOERR
#endif

int main()
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
#ifdef NOERR
	freopen("err.txt", "w", stderr);
#endif
#else
	//freopen("sum.in", "r", stdin);
	//freopen("sum.out", "w", stdout);
#endif

#ifdef ERR_CATCH
	try
	{
#endif

#ifdef ST
		ios_base::sync_with_stdio(false);
		while (true)
			solve();
#endif

#ifdef CASES
#define MULTITEST
#endif

#ifdef MULTITEST
		int t;
		scanf("%d", &t);
		char endl1[5];
		fgets(endl1, 5, stdin);
		for (int i = 0; i < t; ++i)
		{
#ifdef CASES
			printf("Case #%d: ", i + 1);
#endif
			solve();
#ifdef CASES
			eprintf("Passed case #%d:\n", i + 1);
#endif
		}
#else
		solve();
#endif

#ifdef ERR_CATCH
	}
	catch (logic_error e)
	{
		dbg(e.what());
		puts("___________________________________________________________________________");
		exit(0);
	}
#endif
	dbg_time();
}