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 <int MOD_>
  struct modnum {
    static constexpr int MOD = MOD_;
    static_assert(MOD_ > 0, "MOD must be positive");
  
   private:
    using ll = long long;
    int v;
    static int minv(int a, int m) {
      a %= m;
      assert(a);
      return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
    }
  
   public:
    modnum() : v(0) {}
    modnum(ll v_) : v(int(v_ % MOD)) {
      if (v < 0) v += MOD;
    }
    explicit operator int() const { return v; }
    friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
      return out << int(n);
    }
    friend std::istream& operator>>(std::istream& in, modnum& n) {
      ll v_;
      in >> v_;
      n = modnum(v_);
      return in;
    }
    friend bool operator==(const modnum& a, const modnum& b) {
      return a.v == b.v;
    }
    friend bool operator!=(const modnum& a, const modnum& b) {
      return a.v != b.v;
    }
    modnum inv() const {
      modnum res;
      res.v = minv(v, MOD);
      return res;
    }
    friend modnum inv(const modnum& m) { return m.inv(); }
    modnum neg() const {
      modnum res;
      res.v = v ? MOD - v : 0;
      return res;
    }
    friend modnum neg(const modnum& m) { return m.neg(); }
    modnum operator-() const { return neg(); }
    modnum operator+() const { return modnum(*this); }
    modnum& operator++() {
      v++;
      if (v == MOD) v = 0;
      return *this;
    }
    modnum& operator--() {
      if (v == 0) v = MOD;
      v--;
      return *this;
    }
    modnum& operator+=(const modnum& o) {
      v += o.v;
      if (v >= MOD) v -= MOD;
      return *this;
    }
    modnum& operator-=(const modnum& o) {
      v -= o.v;
      if (v < 0) v += MOD;
      return *this;
    }
    modnum& operator*=(const modnum& o) {
      v = int(ll(v) * ll(o.v) % MOD);
      return *this;
    }
    modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
    friend modnum operator++(modnum& a, int) {
      modnum r = a;
      ++a;
      return r;
    }
    friend modnum operator--(modnum& a, int) {
      modnum r = a;
      --a;
      return r;
    }
    friend modnum operator+(const modnum& a, const modnum& b) {
      return modnum(a) += b;
    }
    friend modnum operator-(const modnum& a, const modnum& b) {
      return modnum(a) -= b;
    }
    friend modnum operator*(const modnum& a, const modnum& b) {
      return modnum(a) *= b;
    }
    friend modnum operator/(const modnum& a, const modnum& b) {
      return modnum(a) /= b;
    }
  };
  template <typename T>
  T pow(T a, long long b) {
    assert(b >= 0);
    T r = 1;
    while (b) {
      if (b & 1) r *= a;
      b >>= 1;
      a *= a;
    }
    return r;
  }
  using num = modnum<998244353>;
  vector<num> fact, ifact;
  void init() {
    int N = 1100000;
    fact = {1};
    for (int i = 1; i < N; i++) fact.push_back(i * fact[i - 1]);
    ifact.resize(N);
    ifact.back() = 1 / fact.back();
    for (int i = N - 1; i > 0; i--) ifact[i - 1] = i * ifact[i];
  }
  num ncr(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    init();
    int n;
    cin >> n;
    vector<int> x(n), y(n), s(n);
    for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> s[i];
    num ans = x.back() + 1;
    vector<num> subt(n, 0);
    num cur = 1;
    for (int i = n - 1; i >= 0; i--) {
      num go = cur - (1 - s[i]);
      ans += go * (x[i] - y[i]);
      cur += go;
      int st = -1;
      int en = n;
      while (st + 1 < en) {
        int mi = (st + en) / 2;
        if (x[mi] > y[i]) {
          en = mi;
        } else {
          st = mi;
        }
      }
      assert(en <= i);
      subt[en] += go;
      cur -= subt[i];
    }
    cout << ans << '\n';
  }
  