Problem:
Score : 500 points

Problem Statement

You have a string S of length N.
Initially, all characters in S are 1s.

You will perform queries Q times.
In the i-th query, you are given two integers L_i, R_i and a character D_i (which is a digit).
Then, you must replace all characters from the L_i-th to the R_i-th (inclusive) with D_i.

After each query, read the string S as a decimal integer, and print its value modulo 998,244,353.

Constraints

1 \leq N, Q \leq 200,000

1 \leq L_i \leq R_i \leq N

1 \leq D_i \leq 9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N Q
L_1 R_1 D_1
:
L_Q R_Q D_Q

Output

Print Q lines.
In the i-th line print the value of S after the i-th query, modulo 998,244,353.

Sample Input 1

8 5
3 6 2
1 4 7
3 8 3
2 2 2
4 5 1

Sample Output 1

11222211
77772211
77333333
72333333
72311333

Sample Input 2

200000 1
123 456 7

Sample Output 2

641437905

Don't forget to take the modulo.

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
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; }
using ll = long long;
const long double PI = acos(-1.0L);
ll GCD(ll a, ll b) { return b?GCD(b, a%b):a; }
ll LCM(ll a, ll b) { return a/GCD(a, b)*b; }

template <typename T, typename E>
struct LazeSegTree {
    // 0-indexed
    typedef function<T(T, T)> F;
    typedef function<T(T, E)> G;
    typedef function<E(E, E)> H;
    int n, height;
    F f;    // クエリ処理の関数
    G g;    // 遅延伝搬を反映させる関数
    H h;    // 遅延処理の更新を行う関数
    T ti;   // クエリ処理の単位元
    E ei;   // 遅延配列の単位元
    vector<T> dat;
    vector<E> laz;
    LazeSegTree(){}
    LazeSegTree(F f, G g, H h, T ti, E ei): 
        f(f), g(g), h(h), ti(ti), ei(ei) {}
    
    // 全体を初期化
    void init(int n_) {
        n = 1, height = 0;
        while(n < n_) n <<= 1, ++height;
        dat.assign(n<<1, ti);
        laz.assign(n<<1, ei);
    }

    // 初期配列をセグ木の葉に
    void build(const vector<T> &v) {
        int n_ = v.size();
        init(n_);
        for(int i = 0; i < n_; ++i) dat[n+i] = v[i];
        for(int i = n-1; i; --i) dat[i] = f(dat[(i<<1)|0], dat[(i<<1)|1]);
    }

    // 遅延伝搬の処理を行う
    inline T reflect(int k) {
        return laz[k] == ei ? dat[k] : g(dat[k], laz[k]);
    }

    // 遅延を子ノードに伝搬
    inline void propagate(int k) {
        if(laz[k] == ei) return;
        laz[(k<<1)|0] = h(laz[(k<<1)|0], laz[k]);
        laz[(k<<1)|1] = h(laz[(k<<1)|1], laz[k]);
        dat[k] = reflect(k);
        laz[k] = ei;
    }

    inline void thrust(int k) {
        for(int i = height; i; --i) propagate(k>>i);
        dat[k] = reflect(k);
        laz[k] = ei;
    }

    // クエリ処理をマージ
    inline void recalc(int k) {
        while(k >>= 1) dat[k] = f(reflect((k<<1)|0), reflect((k<<1)|1));
    }

    void update(int a, int b, E x) {
        if(a >= b) return;
        thrust(a += n);
        thrust(b += n-1);
        for(int l = a, r = b+1; l < r; l >>= 1, r >>= 1) {
            if(l&1) laz[l] = h(laz[l], x), l++;
            if(r&1) --r, laz[r] = h(laz[r], x);
        }
        recalc(a);
        recalc(b);
    }

    // buildをバラバラに行う
    void set_val(int a, T x) {
        thrust(a += n);
        dat[a] = x; laz[a] = ei;
        recalc(a);
    }

    T query(int a, int b) {
        if(a >= b) return ti;
        thrust(a += n);
        thrust(b += n-1);
        T vl = ti, vr = ti;
        for(int l = a, r = b+1; l < r; l >>= 1, r >>= 1) {
            if(l&1) vl = f(vl, reflect(l++));
            if(r&1) vr = f(reflect(--r), vr);
        }
        return f(vl, vr);
    }

    template <typename C>
    int find(int st, C &check, T &acc, int k, int l, int r) {
        if(l+1 == r) {
            acc = f(acc, reflect(k));
            return check(acc) ? k-n : -1;
        }
        propagate(k);
        int m = (l+r)>>1;
        if(m <= st) return find(st, check, acc, (k<<1)|1, m, r);
        if(st <= 1 && !check(f(acc, dat[k]))) {
            acc = f(acc, dat[k]);
            return -1;
        }
        int vl = find(st, check, acc, (k<<1)|0, l, m);
        if(~vl) return vl;
        return find(st, check, acc, (k<<1)|1, m, r);
    }

    template <typename C>
    int find(int st, C &check) {
        T acc = ti;
        return find(st, check, acc, 1, 0, n);
    }
};

// auto mod int
// const int mod = 1000000007;
const int mod = 998244353;
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;}

    bool operator==(const mint a) { return x == a.x; }
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

using P = pair<mint, mint>;


int main() {
    int n, q; cin >> n >> q;
    auto f = [](P l, P r) { return P(l.first*r.second+r.first, l.second*r.second); };
    auto g = [](P l, mint r) { return P((l.second-1)/9 * r, l.second); };
    auto h = [](mint l, mint r) { return r; };

    LazeSegTree<P, mint> lseg(f, g, h, P(0, 1), (mint)0);
    lseg.build(vector<P>(n, P(1, 10)));

    for(int i = 0; i < q; ++i) {
        int cl, cr, d; cin >> cl >> cr >> d;
        cl--; cr--;
        lseg.update(cl, cr+1, d);
        cout << lseg.query(0, n).first << endl;
    }
}