#include using namespace std; #define rep(i,n) for(ll i=0;i=0;i--) #define perl(i,r,l) for(ll i=(r)-1;i>=l;i--) #define fi first #define se second #define pb push_back #define ins insert #define pqueue(x) priority_queue,greater> #define all(x) (x).begin(),(x).end() #define CST(x) cout<; using vvl=vector>; using pl=pair; using vpl=vector; using vvpl=vector; const ll MOD=1000000007; const ll MOD9=998244353; const int inf=1e9+10; const ll INF=4e18; const ll dy[9]={1,0,-1,0,1,1,-1,-1,0}; const ll dx[9]={0,-1,0,1,1,-1,1,-1,0}; template inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; } template inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; } const int mod = MOD9; const int max_n = 200005; 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; } bool operator==(const mint &p) const { return x == p.x; } bool operator!=(const mint &p) const { return x != p.x; } // 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[40][1024][1024];//j=末尾、k=xor; int main(){ rep(i,40)rep(j,1024)rep(k,1024)dp[i][j][k]=0; ll n,k,x,y;cin >> n >> k >> x >> y; set st;rep(i,k){ ll p;cin >> p;st.insert(p); } for(auto p:st)dp[0][p][p]=1; rep(i,n-1){ rep(j,1024){ rep(k,1024){ for(auto p:st){ if(j==p)continue; dp[i+1][p][k^p]+=dp[i][j][k]; } } } } mint ans=0; rep(j,1024)rep(k,1024){ if(x<=k&&k<=y)ans+=dp[n-1][j][k]; } cout << ans <