#include <bits/stdc++.h>

#pragma GCC optimize(3, "Ofast", "inline")

using namespace std;
typedef long long ll;

template<const int Mod>
struct ntt {
    inline void add(int &a, int b) {
        a += b;
        if (a >= Mod) a -= Mod;
    }

    inline void sub(int &a, int b) {
        a -= b;
        if (a < 0) a += Mod;
    }

    inline int add2(int a, int b) {
        a += b;
        if (a >= Mod) a -= Mod;
        return a;
    }

    inline int sub2(int a, int b) {
        a -= b;
        if (a < 0) a += Mod;
        return a;
    }

    inline int mul(int a, int b) { return (1ll * a * b % Mod); }

    inline int powmod(int a, long long b) {
        int res = 1;
        while (b > 0) {
            if (b & 1) res = mul(res, a);
            a = mul(a, a);
            b >>= 1;
        }
        return res;
    }

    inline int inv(int a) {
        a %= Mod;
        if (a < 0) a += Mod;
        int b = Mod, u = 0, v = 1;
        while (a) {
            int t = b / a;
            b -= t * a;
            swap(a, b);
            u -= t * v;
            swap(u, v);
        }
        assert(b == 1);
        if (u < 0) u += Mod;
        return u;
    }

    int limit, root;
    vector<int> A, B;

    ntt() {
        int tmp = Mod - 1;
        limit = 0;
        while (tmp % 2 == 0) {
            tmp /= 2;
            limit++;
        }
        root = 2;
        while (powmod(root, (Mod - 1) >> 1) == 1) root++;
        A.resize(limit);
        B.resize(limit);
        for (int i = 0; i < limit; ++i) {
            sub(A[i], powmod(root, (Mod - 1) >> (i + 2)));
            B[i] = inv(A[i]);
        }
    }

    void fft(vector<int> &a, bool inv) {
        const int n = a.size();
        assert((n & (n - 1)) == 0);
        assert(__builtin_ctz(n) <= limit);
        if (!inv) {
            for (int m = n; m >>= 1;) {
                int w = 1;
                for (int s = 0, k = 0; s < n; s += 2 * m) {
                    for (int i = s, j = s + m; i < s + m; ++i, ++j) {
                        int x = a[i], y = mul(a[j], w);
                        a[j] = (x >= y ? x - y : x + Mod - y);
                        a[i] = (x + y >= Mod ? x + y - Mod : x + y);
                    }
                    w = mul(w, A[__builtin_ctz(++k)]);
                }
            }
        } else {
            for (int m = 1; m < n; m *= 2) {
                int w = 1;
                for (int s = 0, k = 0; s < n; s += 2 * m) {
                    for (int i = s, j = s + m; i < s + m; ++i, ++j) {
                        int x = a[i], y = a[j];
                        a[j] = (x >= y ? x - y : x + Mod - y);
                        a[j] = mul(a[j], w);
                        a[i] = (x + y >= Mod ? x + y - Mod : x + y);
                    }
                    w = mul(w, B[__builtin_ctz(++k)]);
                }
            }
        }
    }

    vector<int> multiply(vector<int> a, vector<int> b, int eq = 0) {
        int need = a.size() + b.size() - 1;
        int nbase = 0;
        while ((1 << nbase) < need) nbase++;
        int sz = 1 << nbase;
        a.resize(sz);
        b.resize(sz);
        fft(a, 0);
        if (eq) b = a; else fft(b, 0);
        int inv_sz = inv(sz);
        for (int i = 0; i < sz; i++) {
            a[i] = mul(mul(a[i], b[i]), inv_sz);
        }
        fft(a, 1);
        a.resize(need);
        return a;
    }

    vector<int> square(vector<int> a) {
        return multiply(a, a, 1);
    }
};

const int mod = 998244353;

int x[4];

int k, p;
#define Poly vector<int>

inline int powmod(int a, long long b) {
    int res = 1;
    while (b > 0) {
        if (b & 1) res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        b >>= 1;
    }
    return res;
}

void solve() {
    for (int i = 1; i <= 3; i++) cin >> x[i];
    cin >> k >> p;
    ntt<mod> t;// mod为对应的模数
    Poly A, res;
    A.resize(3);
    A[0] = A[1] = A[2] = powmod(3, mod - 2);
    res.resize(1);
    res[0] = 1;
    int b = k;
    while (b) {
        if (b & 1) res = t.multiply(res, A);
        b >>= 1;
        A = t.multiply(A, A);
    }
    ll L1 = x[1] - x[2], L2 = x[1] - x[3], L3 = x[2] - x[3];
    ll Ans = 0;
    for (int i = 0; i < res.size(); i++) {
        ll d = i - k;
        Ans = (Ans + (abs(L1 + d) + abs(L2 + d) + abs(L3 + d)) % mod * res[i] % mod) % mod;
    }
    cout << Ans * powmod(2, mod - 2) % mod << "\n";
    // max( a , b , c ) - min( a , b , c ) = 0.5 * ( | a - b | + | b - c | + | c - a |)
    // 每一次我会随机在（a,b,c） 中选择一个
    // 有 p 的概率 + 1
    // 有 1-p 的概率 - 1
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int _ = 1;
//    cin >> _;
    for (int o = 1; o <= _; o++) {
//      cout << "Case #<<" << o << ": ";
        solve();
    }
    return 0;
}

/*




*/