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("O3")
  #pragma GCC optimize("O3")
  #pragma GCC target("avx2")
  using namespace std;
  int main() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    long long n;
    cin >> n;
    vector<long long> X(n), Y(n);
    unordered_map<long long, long long> vals;
    vector<long long> V(2 * n + 1);
    long long cur = 0;
    vector<long long> S(2 * n + 1), T(2 * n + 1, 0);
    vector<bool> s(n);
    for (int i = (0); i <= ((int)(n)-1); ++i) {
      long long k;
      cin >> X[i] >> Y[i] >> k;
      if (k == 1)
        s[i] = true;
      else
        s[i] = false;
      V[2 * i] = X[i];
      V[2 * i + 1] = Y[i];
    }
    V[2 * n] = 0;
    sort(V.begin(), V.end());
    for (int i = (0); i <= ((int)(2 * n + 1) - 1); ++i) {
      vals[V[i]] = i;
    }
    S[0] = 0;
    for (int i = (0); i <= ((int)(n)-1); ++i) {
      while (cur < vals[X[i]]) {
        S[cur + 1] = S[cur];
        ++cur;
      }
      T[i] = ((X[i] - Y[i] + S[vals[X[i]]] - S[vals[Y[i]]]) % 998244353 +
              998244353) %
             998244353;
      S[cur] = (S[cur] + T[i]) % 998244353;
    }
    long long res = (X[n - 1] + 1) % 998244353;
    for (int i = (0); i <= ((int)(n)-1); ++i) {
      if (s[i]) {
        res = (res + T[i]) % 998244353;
      }
    }
    cout << (res + 998244353) % 998244353 << endl;
    return 0;
  }
  