#include <bits/stdc++.h>

using namespace std;

#define all(hoge) (hoge).begin(), (hoge).end()
#define en '\n'
using ll = long long;
using ull = unsigned long long;
#define rep(i, m, n) for(ll i = (ll)(m); i < (ll)(n); ++i)
#define rep2(i, m, n) for(ll i = (ll)(n)-1; i >= (ll)(m); --i)
#define REP(i, n) rep(i, 0, n)
#define REP2(i, n) rep2(i, 0, n)
template<class T> using vec = vector<T>;
template<class T> using vvec = vector<vec<T>>;
typedef pair<ll, ll> P;
using tp = tuple<ll, ll, ll>;

constexpr long long INF = 1LL << 60;
constexpr int INF_INT = 1 << 25;
//constexpr long long MOD = (ll) 1e9 + 7;
constexpr long long MOD = 998244353LL;
using ld = long double;
static const ld pi = 3.141592653589793L;

using Array = vector<ll>;
using Matrix = vector<Array>;

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

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}


void solve() {
    ll n, k, x, y;
    cin >> n >> k >> x >> y;
    vec<ll> a(k);
    REP(i, k) {
        cin >> a[i];
    }

    //重複を取り除く
    sort(all(a));
    auto result = unique(all(a));
    a.erase(result, a.end());

    vvec<ll> dp(n + 1, vec<ll>(1024, -1));
    auto dfs = [&](auto &&self, ll i, ll j) -> ll {
        if (dp[i][j] !=-1) return dp[i][j];
        if (i == 0) return dp[i][j] = j==0;
        if (i == 1) {
            ll ret = 0;
            REP(l, a.size()) {
                ret += self(self, i - 1, (j ^ a[l]));
            }
            return dp[i][j] = ret;
        }
        //２連続でa[l]をつかうと同じ合計xorになるので、除く。
        if (i == 2) {
            ll ret = 0;
            REP(l, a.size()) {
                ret += self(self, i - 1, (j ^ a[l]));
            }
            ret -= self(self, i - 2, j) * k;
            return dp[i][j] = ret;
        }
        ll ret = 0;
        REP(l, a.size()) {
            ret += self(self, i - 1, (j ^ a[l]));
        }
        ret -= self(self, i - 2, j) * (k - 1);
        return dp[i][j] = ret;
    };
    ll ans = 0;
    chmin(y, 1023LL);
    rep(i, x, y + 1) {
        ans += dfs(dfs, n, i);
        ans %= MOD;
    }
    cout << ans << en;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
/*
    ll t;
    cin >> t;
    REP(i, t - 1) {
        solve();
    }*/

    solve();

    return 0;
}