#pragma GCC target("avx") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #include // #include // #include // #include // using namespace __gnu_pbds; // #include // namespace multiprecisioninteger = boost::multiprecision; // using cint=multiprecisioninteger::cpp_int; using namespace std; using ll=long long; #define double long double using datas=pair; using ddatas=pair; using tdata=pair; using vec=vector; using mat=vector; using pvec=vector; using pmat=vector; // using llset=tree,rb_tree_tag,tree_order_statistics_node_update>; #define For(i,a,b) for(i=a;i<(ll)b;++i) #define bFor(i,b,a) for(i=b,--i;i>=(ll)a;--i) #define rep(i,N) For(i,0,N) #define rep1(i,N) For(i,1,N) #define brep(i,N) bFor(i,N,0) #define brep1(i,N) bFor(i,N,1) #define all(v) (v).begin(),(v).end() #define allr(v) (v).rbegin(),(v).rend() #define vsort(v) sort(all(v)) #define vrsort(v) sort(allr(v)) #define endl "\n" #define eb emplace_back #define print(x) cout< ostream& operator<<(ostream& os,pair& p){return os<<"{"< inline bool chmax(T& a,T b){bool x=a inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;} void startupcpp(){ cin.tie(0); ios::sync_with_stdio(false); cout<0){ if(n&1)res=res*a%m; a=a*a%m; n>>=1; } return res; } ll gcd(ll a,ll b){if(!b)return abs(a);return (a%b==0)?abs(b):gcd(b,a%b);} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll countdigits(ll n){ ll ans=0; while(n){n/=10;ans++;} return ans; } ll sumdigits(ll n){ ll ans=0; while(n){ans+=n%10;n/=10;} return ans; } ll ans; int main(){ // startupcpp(); // int codeforces;cin>>codeforces;while(codeforces--){ ll i,j,k,N,X,K,cal; cin>>N>>X; K=X>>9; vec a(N),sz(1<<9,0); mat g(1<<9,vec(1<<9,0)),cnt(1<<9,vec(9,0)); rep(i,N){ cin>>a[i]; ++sz[a[i]>>9]; ++g[a[i]>>9][a[i]&511]; rep(j,9)cnt[a[i]>>9][j]+=a[i]>>j&1; } if(!K){ rep(i,1<<9){ rep(j,1<<9){ ans+=(i<<9|j)*g[i][j]*(g[i][j]-1)/2; For(k,j+1,1<<9){ if((j^k)>=(X&511))continue; ans+=(i<<9|(j|k))*g[i][j]*g[i][k]; } } } print(ans); return 0; } rep(i,1<<9){ cal=sz[i]*(sz[i]-1)/2; ans+=cal*i<<9; rep(k,9){ cal=cnt[i][k]*(cnt[i][k]-1)/2+cnt[i][k]*(sz[i]-cnt[i][k]); ans+=cal<=(X&511))continue; ans+=((i|(i^K))<<9|(j|k))*g[i][j]*g[i^K][k]; } } } For(j,i+1,1<<9){ if((i^j)>=K)continue; else if((i^j)