#include <bits/stdc++.h>
using namespace std;
#define repd(i,a,b) for (ll i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define all(x) (x).begin(),(x).end()
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; }
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
using Graph = vector<vector<ll>>;
const long long INF = 1LL<<60;
const long long MOD = 1000000007;

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 998244353;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

mint dp[32][1025];

int main()
{  
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n,k,x,y;
    cin>>n>>k>>x>>y;
    vec a(k);
    rep(i,k)cin>>a[i];
    dp[0][0]=1;
    rep(i,n){
        rep(j,1024){
            if(i==0)dp[i+2][j]-=dp[i][j]*k;
            else dp[i+2][j]-=dp[i][j]*(k-1);
            rep(l,k){
                dp[i+1][j^a[l]]+=dp[i][j];
            }
        }
    }
    mint ans=0;
    repd(i,x,y+1){
        ans+=dp[n][i];
    }
    cout<<ans<<endl;
    return 0;
}