思路： 根据异或的性质，可以把 ∑ l r i ⊕ x \sum_l^r i \oplus x ∑lr​i⊕x划分成一些连续的区间，对每个区间计算答案。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
#define fi first
#define se second
#define pb push_back
int a[N],n,q;
int b[N];
LL sum[N];
int len;
LL qc[N];
const LL mod =998244353;
void init(){
  sort(a+1,a+1+n);
  for(int i=1;i<=n;i++)b[i]=a[i];
  len=unique(b+1,b+1+n)-b-1;
  for(int i=1;i<=n;i++)a[i]=lower_bound(b+1,b+1+len,a[i])-b;
  for(int i=1;i<=n;i++){
    sum[a[i]]++;
  }
  for(int i=1;i<len;i++){
    sum[i]=(0ll+sum[i-1]+sum[i])%mod;
    qc[i]=(1ll*sum[i]*sum[i]%mod*(b[i+1]-b[i])%mod+qc[i-1])%mod;
  }
  sum[len]+=sum[len-1];
  sum[len]%=mod;
}
LL f(int r){
  int x=upper_bound(b+1,b+1+len,r)-b-1;
  if(!x)return 0;
  LL res=qc[x-1];
  return (res+sum[x]*sum[x]%mod*(r-b[x]+1)%mod)%mod;
}
LL g(int l,int r){
  return (f(r)-f(l-1)+mod)%mod;
}
LL get(int r,int x){
  if(r<0)return 0;
  LL tot=0;
  int res=0;
  for(int i=30;~i;i--){
    if(r&(1<<i)){
        if(!(x&(1<<i))){
          tot+=g(res,res+(1<<i)-1);
          res|=(1<<i);
        }else{
          tot+=g(res|(1<<i),(res|(1<<i))+(1<<i)-1);
        }
    }else{
      if((1<<i)&x)res|=(1<<i);
    }
    tot%=mod;
  }
  tot+=g(res,res);
  tot%=mod;
  return tot;
}
int main() {
  ios::sync_with_stdio(false);
  cin>>n>>q;
  for(int i=1;i<=n;i++)cin>>a[i];
  init();
  for(int i=1;i<=q;i++){
    int l,r,x;
    cin>>l>>r>>x;
    cout<<(get(r,x)-get(l-1,x)+mod)%mod<<'\n';
  }
  return 0;
}