description: An ant moves on the real line with constant speed of 1 unit per second. It starts at 0 and always moves to the right (so its position increases by 1 each second).
  
  There are n portals, the i-th of which is located at position x_i and teleports to position y_i < x_i. Each portal can be either active or inactive. The initial state of the i-th portal is determined by s_i: if s_i=0 then the i-th portal is initially inactive, if s_i=1 then the i-th portal is initially active. When the ant travels through a portal (i.e., when its position coincides with the position of a portal): 
  
    * if the portal is inactive, it becomes active (in this case the path of the ant is not affected); 
    * if the portal is active, it becomes inactive and the ant is instantly teleported to the position y_i, where it keeps on moving as normal. 
  
  
  
  How long (from the instant it starts moving) does it take for the ant to reach the position x_n + 1? It can be shown that this happens in a finite amount of time. Since the answer may be very large, compute it modulo 998 244 353.
  
  Input
  
  The first line contains the integer n (1≤ n≤ 2⋅ 10^5) — the number of portals.
  
  The i-th of the next n lines contains three integers x_i, y_i and s_i (1≤ y_i < x_i≤ 10^9, s_i∈\{0,1\}) — the position of the i-th portal, the position where the ant is teleported when it travels through the i-th portal (if it is active), and the initial state of the i-th portal.
  
  The positions of the portals are strictly increasing, that is x_1<x_2<⋅⋅⋅<x_n. It is guaranteed that the 2n integers x_1,   x_2,   ...,   x_n,   y_1,   y_2,   ...,   y_n are all distinct.
  
  Output
  
  Output the amount of time elapsed, in seconds, from the instant the ant starts moving to the instant it reaches the position x_n+1. Since the answer may be very large, output it modulo 998 244 353.
  
  Examples
  
  Input
  
  
  4
  3 2 0
  6 5 1
  7 4 0
  8 1 1
  
  
  Output
  
  
  23
  
  
  Input
  
  
  1
  454971987 406874902 1
  
  
  Output
  
  
  503069073
  
  
  Input
  
  
  5
  243385510 42245605 0
  644426565 574769163 0
  708622105 208990040 0
  786625660 616437691 0
  899754846 382774619 0
  
  
  Output
  
  
  899754847
  
  
  Input
  
  
  5
  200000000 100000000 1
  600000000 400000000 0
  800000000 300000000 0
  900000000 700000000 1
  1000000000 500000000 0
  
  
  Output
  
  
  3511295
  
  Note
  
  Explanation of the first sample: The ant moves as follows (a curvy arrow denotes a teleporting, a straight arrow denotes normal movement with speed 1 and the time spent during the movement is written above the arrow). $$$ 0 \stackrel{6}{\longrightarrow} 6 \leadsto 5 \stackrel{3}{\longrightarrow} 8 \leadsto 1 \stackrel{2}{\longrightarrow} 3 \leadsto 2 \stackrel{4}{\longrightarrow} 6 \leadsto 5 \stackrel{2}{\longrightarrow} 7 \leadsto 4 \stackrel{2}{\longrightarrow} 6 \leadsto 5 \stackrel{4}{\longrightarrow} 9  Notice that the total time is 6+3+2+4+2+2+4=23$$$.
  
  Explanation of the second sample: The ant moves as follows (a curvy arrow denotes a teleporting, a straight arrow denotes normal movement with speed 1 and the time spent during the movement is written above the arrow). $$$ 0 \stackrel{454971987}{\longrightarrow} 454971987 \leadsto 406874902 \stackrel{48097086}{\longrightarrow} 454971988  Notice that the total time is 454971987+48097086=503069073$$$.
  
  Explanation of the third sample: Since all portals are initially off, the ant will not be teleported and will go straight from 0 to x_n+1=899754846+1=899754847.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <const int32_t MOD>
  struct ModInt {
    int32_t x;
    ModInt() : x(0) {}
    ModInt(long long u) : x(u % MOD) {
      x %= MOD;
      if (x < 0) x += MOD;
    }
    friend bool operator==(const ModInt &a, const ModInt &b) {
      return a.x == b.x;
    }
    friend bool operator!=(const ModInt &a, const ModInt &b) {
      return a.x != b.x;
    }
    friend bool operator<(const ModInt &a, const ModInt &b) { return a.x < b.x; }
    friend bool operator>(const ModInt &a, const ModInt &b) { return a.x > b.x; }
    friend bool operator<=(const ModInt &a, const ModInt &b) {
      return a.x <= b.x;
    }
    friend bool operator>=(const ModInt &a, const ModInt &b) {
      return a.x >= b.x;
    }
    static ModInt sign(long long k) {
      return ((k & 1) ? ModInt(MOD - 1) : ModInt(1));
    }
    ModInt &operator+=(const ModInt &m) {
      x += m.x;
      if (x >= MOD) x -= MOD;
      return *this;
    }
    ModInt &operator-=(const ModInt &m) {
      x -= m.x;
      if (x < 0LL) x += MOD;
      return *this;
    }
    ModInt &operator*=(const ModInt &m) {
      x = (1LL * x * m.x) % MOD;
      return *this;
    }
    friend ModInt operator-(const ModInt &a) {
      ModInt res(a);
      if (res.x) res.x = MOD - res.x;
      return res;
    }
    friend ModInt operator+(const ModInt &a, const ModInt &b) {
      return ModInt(a) += ModInt(b);
    }
    friend ModInt operator-(const ModInt &a, const ModInt &b) {
      return ModInt(a) -= ModInt(b);
    }
    friend ModInt operator*(const ModInt &a, const ModInt &b) {
      return ModInt(a) *= ModInt(b);
    }
    static long long fp(long long u, long long k) {
      long long res = 1LL;
      while (k > 0LL) {
        if (k & 1LL) res = (res * u) % MOD;
        u = (u * u) % MOD;
        k /= 2LL;
      }
      return res;
    }
    static constexpr long long mod() { return MOD; }
    ModInt fastpow(long long k) { return ModInt(fp(x, k)); }
    ModInt inv() {
      assert(x);
      return ModInt(fp(x, MOD - 2));
    }
    ModInt &operator/=(const ModInt &m) { return *this *= ModInt(m).inv(); }
    friend ModInt operator/(const ModInt &a, const ModInt &b) {
      return ModInt(a) *= ModInt(b).inv();
    }
    friend ostream &operator<<(ostream &out, const ModInt &a) {
      return out << a.x;
    }
    friend istream &operator>>(istream &in, ModInt &a) { return in >> a.x; }
  };
  const long long MOD = 998244353;
  using Mint = ModInt<MOD>;
  template <class T>
  class Fenwick {
   public:
    long long lim;
    vector<T> bit;
    Fenwick(long long n) : lim(n + 1), bit(lim) {}
    void upd(long long pos, T val) {
      for (pos++; pos < lim; pos += pos & -pos) bit[pos] += val;
    }
    T sum(long long r) {
      T ret = 0;
      for (; r; r -= r & -r) ret += bit[r];
      return ret;
    }
    T sum(long long l, long long r) { return sum(r) - sum(l); }
  };
  signed main(void) {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    long long nbPortails;
    cin >> nbPortails;
    vector<long long> depart(nbPortails), arrive(nbPortails), state(nbPortails);
    vector<Mint> cost(nbPortails);
    for (long long i = 0; i < nbPortails; ++i)
      cin >> depart[i] >> arrive[i] >> state[i];
    Fenwick<Mint> fen(nbPortails);
    for (long long iPortail = 0; iPortail < nbPortails; ++iPortail) {
      long long fst =
          lower_bound(depart.begin(), depart.end(), arrive[iPortail]) -
          depart.begin();
      cost[iPortail] =
          depart[iPortail] - arrive[iPortail] + fen.sum(fst, iPortail);
      fen.upd(iPortail, cost[iPortail]);
    }
    Mint ret = depart.back() + 1;
    for (long long i = 0; i < nbPortails; ++i)
      if (state[i]) ret += cost[i];
    cout << ret << endl;
  }
  