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)

/*
 * じょえチャンネル
 * 高評価・チャンネル登録よろしくおねがいします！
 * https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
 */

#include <bits/stdc++.h>

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)

//here!!!
// define int long long !!!!!

#define int long long

// define int long long !!!!!


//#define mod (int)((1e9)+7)
constexpr int mod = 998244353ll;

#define intt long long
#define itn long long
#define P pair<int,int>

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)

#define ALL(v) v.begin(),v.end()

using namespace std;

//Library
//モッドパウ
inline int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
            res %= m;
        }
        x = x * x % m;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}


//saidai-kouyakusuu
inline int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}


//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

inline int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}


template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
    if (lhs < rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
    if (lhs > rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
void Yes(){
    cout<<"Yes"<<endl;
}
void No(){
    cout<<"No"<<endl;
}
void YES(){
    cout<<"YES"<<endl;
}
void NO(){
    cout<<"NO"<<endl;
}


#define fin(i) scanf("%lld",&i)
#define fout(i) printf("%lld",i)
#define fendl printf("\n")

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    //    cout<<kai(x, y)<<' '<<modpow(kai(y, y), mod - 2)<<endl;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}
//Library-End


#define vecin(v) for(int i=0;i<v.size();i++)scanf("%lld",&v[i]);
#define vecout(v) {for(int i=0;i<(int)v.size();i++)printf("%lld ",v[i]);printf("\n");}

//SegTree
template <class T>
class SegTree {
    int n;                       // 葉の数
    vector<T> data;              // データを格納するvector
    T def;                       // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;    // 点更新で使う処理
    
    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }
    
public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
            function<T(T, T)> _update)
    : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }
    
    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }
    
    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }
    
    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }

class Union_Find {
    vector<int> par;
    vector<int> rankmy;
    vector<int> ookisa;
    
public:
    Union_Find(int size) {
        par = vector<int>(size);
        rankmy = vector<int>(size);
        ookisa=vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
            ookisa[i]=1;
        }
    }
    
    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }
    
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rankmy[x] < rankmy[y]) {
            par[x] = y;
            ookisa[y]+=ookisa[x];
            ookisa[x]=0;
        }
        else {
            par[y] = x;
            ookisa[x]+=ookisa[y];
            ookisa[y]=0;
            if (rankmy[x] == rankmy[y]) {
                rankmy[x]++;
            }
        }
    }
    int size(int i){
        i=find(i);
        return ookisa[i];
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
    
};

//Union-Find-End
#define endl "\n" //interactive の時に注意！！！
int perm[1000005];
void init_perm() {
    perm[0] = 1;
    REP(i, 1000000) {
        perm[i] = perm[i - 1] * i % mod;
    }
}

int nCk(int x, int y) {
    return perm[x] * modpow(perm[x - y], mod - 2) % mod * modpow(perm[y], mod - 2) % mod;
}

double kyori(pair<int, int> f, pair<int, int> s) {
    double ans = 0;
    double t = fabs(f.first - s.first);
    double y = fabs(f.second - s.second);
    ans = sqrt(t * t + y * y);
    return ans;
}


int n,a,b,K,ans;
signed main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cin>>n>>a>>b>>K;
    init_perm();
    if (a<b) {
        swap(a, b);
    }
    rep(i,n+1){
        int k=i*a;
        if (k>K) {
            continue;
        }
        if ((K-k)%(b)) {
            continue;
        }
        int j=(K-k)/(b);
        if (j>n) {
            continue;
        }
        ans+=nCk(n,j)*nCk(n,i)%mod;
        ans%=mod;
    }
    cout<<ans<<endl;
    
}