Problem:
Score : 100 points

Problem Statement

You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types.

0 l r b c: For each i = l, l+1, \dots, {r - 1}, set a_i \gets b \times a_i + c.

1 l r: Print \sum_{i = l}^{r - 1} a_i \bmod 998244353.

Constraints

1 \leq N, Q \leq 500000

0 \leq a_i, c < 998244353

1 \leq b < 998244353

0 \leq l < r \leq N

All values in Input are integer.

Input

Input is given from Standard Input in the following format:

N Q
a_0 a_1 ... a_{N - 1}
\textrm{Query}_0
\textrm{Query}_1
:
\textrm{Query}_{Q - 1}

Output

For each query of the latter type, print the answer.

Sample Input 1

5 7
1 2 3 4 5
1 0 5
0 2 4 100 101
1 0 3
0 1 3 102 103
1 2 5
0 2 5 104 105
1 0 5

Sample Output 1

15
404
41511
4317767

Solution in C++ (GCC 9.2.1 with AC Library v1.1) (judge verdict: Accepted)

#include <bits/stdc++.h>



#include <atcoder/all>
using namespace atcoder;

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#include <boost/multiprecision/cpp_int.hpp>
using namespace std;


using dou =long double;
string yes="yes";
string Yes="Yes";
string YES="YES";
string no="no";
string No="No";
string NO="NO";

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
/*
const ll mod = 998244353ll;
//const ll mod = 10000000000ll;
//const ll mod = 10000;

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, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
*/
#define rep(i, n)         for(ll i = 0; i < (ll)(n); i++)
#define brep(n)           for(int bit=0;bit<(1<<n);bit++)
#define bbrep(n)           for(int bbit=0;bbit<(1<<n);bbit++)
#define erep(i,container) for (auto &i : container)
#define itrep(i,container) for (auto i : container)
#define irep(i, n)        for(ll i = n-1; i >= (ll)0ll; i--)
#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)
#define reprep(i,j,h,w) rep(i,h)rep(j,w)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];
#define pb push_back
#define pf push_front
#define query int qq;std::cin >> qq;rep(qqq,qq)
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define itn int
#define mp make_pair
//#define sum(a) accumulate(all(a),0ll)
#define keta fixed<<setprecision
#define vout(a) erep(qxqxqx,a)std::cout << qxqxqx << ' ';std::cout  << std::endl;
#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))
//#define vvector(name,typ,m,n)vector<vector<typ> > name(m,vector<typ> (n))
#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));
#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));
#define case std::cout <<"Case #" <<qqq+1<<":"
#define RES(a,i,j) a.resize(i);rep(ii,i)a[ii].resize(j);
#define RESRES(a,i,j,k) a.resize(i);rep(ii,i)a[ii].resize(j);reprep(ii,jj,i,j){dp[ii][jj].resize(k)};
#define res resize
#define as assign
#define ffor for(;;)
#define ppri(a,b) std::cout << a<<" "<<b << std::endl
#define pppri(a,b,c) std::cout << a<<" "<<b <<" "<< c<<std::endl
#define ppppri(a,b,c,d) std::cout << a<<" "<<b <<" "<< c<<' '<<d<<std::endl
#define aall(x,n) (x).begin(),(x).begin()+(n)
#define SUM(a) accumulate(all(a),0ll) 
#define stirng string
#define gin(a,b) int a,b;std::cin >> a>>b;a--;b--;
#define popcount __builtin_popcount
#define permu(a) next_permutation(all(a))
//#define grid_input(a,type) int h,w;std::cin >> h>>w;vvector(a,type,h,w,0);reprep(i,j,h,w)std::cin >> a[i][j];

//typedef long long T;
ll ceil(ll a,ll b){
    return ((a+b-1)/b);
}
const int INF = 2000000000;
//const ll INF64 =3223372036854775807ll;
//const ll INF64 = 9223372036854775807ll;
const ll INF64 = 243'000'000'000'000'000'0;

const ll MOD = 998244353ll;
//const ll MOD = 1000003ll;
const ll OD = 1000000000000007ll;
const dou pi=3.141592653589793;
long long modpow(long long a, long long n) { //累乗の余剰
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}


//メモ
//ゲーム(Grundy数とか)の復習をする
//群論の勉強をする？
//学会のスライドを治す(木曜日まで)
//周期性の実験をする
//リスニング力をどうにかする
//マンハッタン距離の問題は45度回転するとうまくいくことがあるらしいよ！
//とりあえずALCはちゃんと埋める(遅延セグ木のやつわかんね～)
//特に理由がなければllを使おうキャンペーン
//

using mint =modint998244353;
struct S{
    mint a;
    int size;
};
struct F{
    mint a,b;
};

S op(S l,S r){
    return S{l.a+r.a ,l.size+r.size};
}
S e(){
    return S{0,0};
}
S mapping(F f,S x){
    return S{x.a*f.a+f.b*x.size, x.size};
}
F composition(F l,F r){
    return F{l.a*r.a, l.a*r.b+l.b};
}
F id(){
    return F{1,0};
}
int main(){
    
std::ios::sync_with_stdio(false); //cin高速化
std::cin.tie(0);
    int n,q;
    std::cin >> n>>q;
    std::vector<S> a(n);
    rep(i,n){
        int d;
        std::cin >> d;
        a[i].a=d;
        a[i].size=1;
    }
    lazy_segtree<S,op,e,F,mapping,composition,id> seg(a);
    //std::cout << a.size() << std::endl;
    rep(qqq,q){
        int t;
        std::cin >> t;
        if(t==1){
            int l,r;
            std::cin >> l>>r;
            std::cout << seg.prod(l,r).a.val() << std::endl;
        }
        else{
            int l,r,b,c;
            std::cin >> l>>r>>b>>c;
            seg.apply(l,r,F{b,c});
        }
    }
}