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;
  long long int min(long long int a, long long int b) { return (a > b) ? b : a; }
  long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
  struct node {
    long long int x;
    node() { x = 0; }
    node operator+(const node& other) {
      node sum = node();
      sum.x = (x + other.x) % 998244353;
      return sum;
    }
  };
  struct SegTree {
    vector<node> t;
    long long int n;
    SegTree(long long int n) {
      this->n = n;
      t.assign(4 * n, node());
    }
    SegTree(vector<long long int>& a) : SegTree(a.size()) { build(a, 1, 1, n); }
    void build(vector<long long int>& a, int v, int tl, int tr) {
      if (tl == tr)
        t[v].x = a[tl - 1];
      else {
        int tm = (tl + tr) / 2;
        build(a, v * 2, tl, tm);
        build(a, v * 2 + 1, tm + 1, tr);
        t[v] = t[v * 2] + t[v * 2 + 1];
      }
    }
    node sum(int v, int tl, int tr, int l, int r) {
      if (l > r) return node();
      if (l == tl && r == tr) return t[v];
      int tm = (tl + tr) / 2;
      return sum(v * 2, tl, tm, l, min(r, tm)) +
             sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
    }
    long long int sum(int l, int r) { return sum(1, 1, n, l, r).x; }
    void update(int v, int tl, int tr, int pos, long long int new_val) {
      if (tl == tr)
        t[v].x = new_val;
      else {
        int tm = (tl + tr) / 2;
        if (pos <= tm)
          update(2 * v, tl, tm, pos, new_val);
        else
          update(2 * v + 1, tm + 1, tr, pos, new_val);
        t[v] = t[2 * v] + t[2 * v + 1];
      }
    }
    void update(int idx, long long int a) { update(1, 1, n, idx, a); }
  };
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int te = 1;
    while (te--) {
      long long int n;
      cin >> n;
      long long int x[n], y[n], f[n];
      for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> f[i];
      long long int ans = (x[n - 1] + 1) % 998244353;
      map<long long int, long long int> m;
      for (int i = 0; i < n; i++) m[x[i]]++, m[y[i]]++;
      int c = 0;
      for (auto i : m) m[i.first] = ++c;
      SegTree ST(c + 5);
      for (int i = 0; i < n; i++) {
        long long int s = ST.sum(m[y[i]], m[x[i]]);
        (s += x[i] - y[i]) %= 998244353;
        ST.update(m[x[i]], s);
        if (f[i]) (ans += s) %= 998244353;
      }
      cout << ans << '\n';
    }
  }
  