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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.*;
import java.util.*;


public class Main {

	private static Scanner sc;
	private static Printer pr;

	private static void solve() {
		final int MOD = 998_244_353;

		int n = sc.nextInt();
		int a = sc.nextInt();
		int b = sc.nextInt();
		long k = sc.nextLong();

		PC pc = new PC(n, MOD);

		long ans = 0;
		for (long aa = 0; aa <= n; aa++) {
			if ((k - aa * a) % b != 0) {
				continue;
			}
			long bb = (k - aa * a) / b;

			for (long c = Math.max(0, aa + bb - n); c <= Math.min(aa, bb); c++) {
//				if (n - aa - bb + c < 0) {
//					continue;
//				}

				long tmp = pc.fact[n];
				tmp *= pc.ifact[(int)(aa - c)];
				tmp %= MOD;
				tmp *= pc.ifact[(int)(bb - c)];
				tmp %= MOD;
				tmp *= pc.ifact[(int)c];
				tmp %= MOD;
				tmp *= pc.ifact[(int)(n - aa - bb + c)];
				tmp %= MOD;

				ans += tmp;
				ans %= MOD;
			}
		}

		pr.println(ans);
	}

    static class PC {
    	// MOD must be a prime number.
    	int MOD;
    	// fact[i] : i! % MOD
    	long[] fact;
    	// ifact[i] : 1/i! % MOD
    	long[] ifact;

    	PC(int size, int MOD) {
    		// O(size)
    		// n=sizeまでのnCrを求める。
    		// nHrはn+r-1Crになってしまうので注意

    		this.MOD = MOD;

    		fact = new long[size + 1];
    		fact[0] = 1;
    		for (int i = 1; i <= size; i++) {
    			fact[i] = fact[i - 1] * i % MOD;
    		}

    		ifact = new long[size + 1];

    		int loop = MOD - 2;
    		long x = fact[size];
    		ifact[size] = 1;
    		while (loop > 0) {
    			if (loop % 2 == 1) {
    				ifact[size] = ifact[size] * x % MOD;
    			}
    			x = x * x % MOD;
    			loop /= 2;
    		}

    		for (int i = size - 1; i >= 0; i--) {
    			ifact[i] = ifact[i + 1] * (i + 1) % MOD;
    		}

    	}

    	// 組合せの数
    	int C(int n, int r) {
    		if (r > n) {
    			return 0;
    		}

    		return (int)(((fact[n] * ifact[n - r]) % MOD) * ifact[r] % MOD);
    	}

    	// 順列
    	int P(int n, int r) {
    		if (r > n) {
    			return 0;
    		}

    		return (int)((fact[n] * ifact[n -r]) % MOD);
    	}

    	// 重複組み合わせ
    	// 異なるn種のものから重複を許してr個を選ぶ場合の数
    	// 0個の種類もあり得る
    	int H(int n, int r) {
    		if (n == 0 && r == 0) {
    			return 1;
    		}

    		return C(n + r - 1, r);
    	}

    	// 組合せの数(nが大きいとき)
    	//   O(r)で求めることができる。rはsizeの大きさまで
    	int C2(long n, int r) {
    		long ret = ifact[r];
    		for (int i = 1; i <= r; i++) {
    			long tmp = (n - r + i) % MOD;
    			ret = (ret * tmp) % MOD;
    		}

    		return (int)ret;
    	}

    	// 第2種スターリング数
    	// n人をちょうどr個のグループに分ける(グループの区別はなし)
    	// グループの区別をする場合はr!S(n,r)。全射の場合の数と同義
    	// O(r log n)
    	int S(long n, int r) {
    		//全射の場合の数を包除原理を使って求めて、1/r!をかける。
    		long ret = 0;
    		for (int i = 1; i <= r; i++) {
    			long tmp = (r - i) % 2 == 0 ? 1 : -1;
    			tmp *= pow(i, n) * C(r, i) % MOD;
    			ret = (ret + tmp + MOD) % MOD;
    		}
    		ret = ret * ifact[r] % MOD;

    		return (int)ret;
    	}

    	long pow(int a, long n) {
    		long loop = n;
    		long ret = 1;
    		long x = a;
    		while (loop > 0) {
    			if (loop % 2 == 1) {
    				ret = ret * x % MOD;
    			}
    			x = x * x % MOD;
    			loop /= 2;
    		}

    		return ret;
    	}

    	// 組合せの数
    	// パスカルの三角形MODなし
		// 限界：n=66 : 66C33=7219428434016265740
    	private final static int LIMIT = 66;
    	private static int to;
    	private static long[][] cache;

    	static long CLong(int n, int r) {
    		if (r > n) {
    			return 0;
    		}

    		if (n > LIMIT) {
    			throw new IllegalArgumentException(Integer.toString(n));
    		}

    		if (cache == null) {
    			cache = new long[LIMIT + 1][];
    			cache[0] = new long[1];
    			cache[0][0] = 1;
    			to = 0;
    		}

    		if (cache[n] == null) {
    			for (int i = to + 1; i <= n; i++) {
    				cache[i] = new long[i + 1];
    				for (int j = 0; j <= i; j++) {
    					if (j == 0 || j == i) {
    						cache[i][j] = 1;
    					} else {
    						if (Long.MAX_VALUE - cache[i - 1][j - 1] < cache[i - 1][j]) {
    			    			throw new IllegalArgumentException("Overflow");
    						} else {
    							cache[i][j] = cache[i - 1][j - 1] + cache[i - 1][j];
    						}
    					}
    				}
    			}
    			to = n;
    		}

    		return cache[n][r];
    	}
    }

	// ---------------------------------------------------
	public static void main(String[] args) {
		sc = new Scanner(INPUT == null ? System.in : new ByteArrayInputStream(INPUT.getBytes()));
		pr = new Printer(System.out);

		solve();

//		pr.close();
		pr.flush();
//		sc.close();
	}

	static String INPUT = null;

	@SuppressWarnings("unused")
	private static class Scanner {
		BufferedReader br;

		Scanner (InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

		private boolean isPrintable(int ch) {
			return ch >= '!' && ch <= '~';
		}

		private boolean isCRLF(int ch) {
			return ch == '\n' || ch == '\r' || ch == -1;
		}

		private int nextPrintable() {
			try {
				int ch;
				while (!isPrintable(ch = br.read())) {
					if (ch == -1) {
						throw new NoSuchElementException();
					}
				}

				return ch;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		String next() {
			try {
				int ch = nextPrintable();
				StringBuilder sb = new StringBuilder();
				do {
					sb.appendCodePoint(ch);
				} while (isPrintable(ch = br.read()));

				return sb.toString();
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		int nextInt() {
			try {
				// parseInt from Integer.parseInt()
				boolean negative = false;
				int res = 0;
				int limit = -Integer.MAX_VALUE;
				int radix = 10;

				int fc = nextPrintable();
				if (fc < '0') {
					if (fc == '-') {
						negative = true;
						limit = Integer.MIN_VALUE;
					} else if (fc != '+') {
						throw new NumberFormatException();
					}
					fc = br.read();
				}
				int multmin = limit / radix;

				int ch = fc;
				do {
					int digit = ch - '0';
					if (digit < 0 || digit >= radix) {
						throw new NumberFormatException();
					}
					if (res < multmin) {
						throw new NumberFormatException();
					}
					res *= radix;
					if (res < limit + digit) {
						throw new NumberFormatException();
					}
					res -= digit;

				} while (isPrintable(ch = br.read()));

				return negative ? res : -res;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		long nextLong() {
			try {
				// parseLong from Long.parseLong()
				boolean negative = false;
				long res = 0;
				long limit = -Long.MAX_VALUE;
				int radix = 10;

				int fc = nextPrintable();
				if (fc < '0') {
					if (fc == '-') {
						negative = true;
						limit = Long.MIN_VALUE;
					} else if (fc != '+') {
						throw new NumberFormatException();
					}
					fc = br.read();
				}
				long multmin = limit / radix;

				int ch = fc;
				do {
					int digit = ch - '0';
					if (digit < 0 || digit >= radix) {
						throw new NumberFormatException();
					}
					if (res < multmin) {
						throw new NumberFormatException();
					}
					res *= radix;
					if (res < limit + digit) {
						throw new NumberFormatException();
					}
					res -= digit;

				} while (isPrintable(ch = br.read()));

				return negative ? res : -res;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		float nextFloat() {
			return Float.parseFloat(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			try {
				int ch;
				while (isCRLF(ch = br.read())) {
					if (ch == -1) {
						throw new NoSuchElementException();
					}
				}
				StringBuilder sb = new StringBuilder();
				do {
					sb.appendCodePoint(ch);
				} while (!isCRLF(ch = br.read()));

				return sb.toString();
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
//				throw new NoSuchElementException();
			}
		}
	}

	private static class Printer extends PrintWriter {
		Printer(OutputStream out) {
			super(out);
		}
	}
}