#include <iostream>
#include <iomanip>
#include <algorithm>
#include <complex>
#include <utility>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <tuple>
#include <cmath>
#include <bitset>
#include <cctype>
#include <set>
#include <map>
#include <unordered_map>
#include <numeric>
#include <functional>
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define PRINT(V) cout << V << "\n"
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())
using namespace std;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline void Yes(bool condition){ if(condition) PRINT("Yes"); else PRINT("No"); }
template<class itr> void cins(itr first,itr last){
    for (auto i = first;i != last;i++){
        cin >> (*i);
    }
}
template<class itr> void array_output(itr start,itr goal){
    string ans = "",k = " ";
    for (auto i = start;i != goal;i++) ans += to_string(*i)+k;
    if (!ans.empty()) ans.pop_back();
    PRINT(ans);
}
ll gcd(ll a, ll b) {
    return a ? gcd(b%a,a) : b;
}
const ll INF = 1e18;
const ll MOD = 1000000007;
const ll MOD2 = 998244353;
typedef pair<ll,ll> P;
typedef pair<double,double> point;
const ll MAX = 200000;
constexpr ll nx[8] = {1,0,-1,0,1,-1,1,-1};
constexpr ll ny[8] = {0,1,0,-1,1,1,-1,-1};
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,q;
    cin >> n >> q;
    vector<ll> a(n),m(n+1,1);
    cins(all(a));
    vector<ll> b = a;
    RSORT(a);
    SORT(b);
    rep(i,n){
        m[i+1] = m[i]*b[i]%MOD2;
    }
    vector<vector<ll>> dp(n+1,vector<ll>(n+1,0));
    dp[0][0] = 1;
    rep(i,n){
        (dp[i+1][0] = dp[i][0]*(a[i]-1))%=MOD2;
        rep(j,1,n+1){
            (dp[i+1][j] = dp[i][j]*(a[i]-1)+dp[i][j-1])%=MOD2;
        }
    }
    ll l,r,p;
    rep(i,q){
        cin >> l >> r >> p;
        ll res = 0;
        rep(j,l,r+1){
            ll k = lower_bound(all(b),j)-b.begin();
            k = n-k;
            if (k < p) continue;
            res = res^(dp[k][p]*m[n-k]%MOD2);
        }
        PRINT(res);
    }   
}