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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx2")
  using namespace std;
  template <class T>
  struct LazySegmentTree : public T {
    using S = typename T::S;
    using F = typename T::F;
    LazySegmentTree(int n) : d(4 * n, T::e()), lz(4 * n, T::id()) {
      for (; sz < n; sz *= 2)
        ;
    }
    void apply(int l, int r, F f) { apply(l, r, f, 0, 0, sz); }
    S prod(int l, int r) { return prod(l, r, 0, 0, sz); }
  
   private:
    int sz = 1;
    vector<S> d;
    vector<F> lz;
    void propagate(int x, int lx, int rx) {
      if (rx - lx == 1) return;
      lz[2 * x + 1] = T::composition(lz[x], lz[2 * x + 1]);
      lz[2 * x + 2] = T::composition(lz[x], lz[2 * x + 2]);
      d[2 * x + 1] = T::mapping(lz[x], d[2 * x + 1], (rx - lx) / 2);
      d[2 * x + 2] = T::mapping(lz[x], d[2 * x + 2], (rx - lx) / 2);
      lz[x] = T::id();
    }
    void apply(int l, int r, F f, int x, int lx, int rx) {
      propagate(x, lx, rx);
      if (rx <= l or r <= lx) return;
      if (l <= lx and rx <= r) {
        lz[x] = T::composition(f, lz[x]);
        d[x] = T::mapping(f, d[x], rx - lx);
        return;
      }
      apply(l, r, f, 2 * x + 1, lx, (lx + rx) / 2);
      apply(l, r, f, 2 * x + 2, (lx + rx) / 2, rx);
      d[x] = T::op(d[2 * x + 1], d[2 * x + 2]);
    }
    S prod(int l, int r, int x, int lx, int rx) {
      propagate(x, lx, rx);
      if (rx <= l or r <= lx) return T::e();
      if (l <= lx and rx <= r) return d[x];
      return T::op(prod(l, r, 2 * x + 1, lx, (lx + rx) / 2),
                   prod(l, r, 2 * x + 2, (lx + rx) / 2, rx));
    }
  };
  const int MOD = 998244353;
  struct Node {
    using S = long long;
    using F = long long;
    S op(S a, S b) { return (a + b) % MOD; }
    S e() { return 0; }
    S mapping(F f, S x, int len) { return (x + f * len) % MOD; }
    F composition(F f, F g) { return (f + g) % MOD; }
    F id() { return 0; }
  };
  struct Portal {
    int x, y;
    bool s;
  };
  void solve() {
    int n;
    cin >> n;
    vector<Portal> portals(n);
    for (auto &portal : portals) cin >> portal.x >> portal.y >> portal.s;
    LazySegmentTree<Node> st(n);
    for (int i = n - 1; i >= 0; --i) {
      if (portals[i].s) {
        st.apply(i, i + 1, 1);
      }
      int lo = -1, hi = i;
      while (lo + 1 < hi) {
        int mid = (lo + hi) / 2;
        if (portals[mid].x >= portals[i].y)
          hi = mid;
        else
          lo = mid;
      }
      st.apply(hi, i, st.prod(i, i + 1));
    }
    long long ans = portals[n - 1].x + 1;
    for (int i = 0; i < n; ++i)
      ans = (ans + st.prod(i, i + 1) * (portals[i].x - portals[i].y)) % MOD;
    cout << ans << "\n";
  }
  signed main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    solve();
  }
  