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: import java.util.*;
  import java.io.*;
  
  public class Main {
      public static void main(String args[]) {
          new Main().run();
      }
  
      FastReader in = new FastReader();
      PrintWriter out = new PrintWriter(System.out);
      void run() {
          work();
          out.flush();
      }
  
      long mod = 998244353;
      long gcd(long a, long b) {
          return a == 0 ? b : gcd(b % a, a);
      }
      long inf = Long.MAX_VALUE / 3;
      void work() {
          int n=ni();
          TreeMap<Integer,Long> tree=new TreeMap<>();
          int[][] A=new int[n+1][3];
          for(int i=1;i<=n;i++){
              A[i]=new int[]{ni(),ni(),ni()};
          }
          long ret=0;
          long cur=1;
          Integer last=A[n][0]+1;
          for(int i=n;i>=0;i--){
              Integer pre=tree.lowerKey(last);
              while(pre!=null&&pre>A[i][0]){
                  long c=tree.get(pre);
                  ret+=cur*(last-pre);
                  ret%=mod;
                  cur-=c;
                  last=pre;
                  pre=tree.lowerKey(last);
              }
              ret=ret+cur*(last-A[i][0]);
              ret%=mod;
              long cur2=cur*2;
              if(A[i][2]==0){
                  cur2--;
              }
              tree.put(A[i][1],cur2-cur);
              last=A[i][0];
              cur=cur2;
              cur%=mod;
          }
          out.println((ret%mod+mod)%mod);
      }
  
      private ArrayList<long[]>[] ngw(int n, int m) {
          ArrayList<long[]>[] graph = (ArrayList<long[]>[]) new ArrayList[n];
          for (int i = 0; i < n; i++) {
              graph[i] = new ArrayList<>();
          }
          for (int i = 1; i <= m; i++) {
              long s = in.nextLong() - 1, e = in.nextLong() - 1, w = in.nextLong();
              graph[(int) s].add(new long[]{e, w});
              graph[(int) e].add(new long[]{s, w});
          }
          return graph;
      }
  
      private ArrayList<Integer>[] ng(int n, int m) {
          ArrayList<Integer>[] graph = (ArrayList<Integer>[]) new ArrayList[n];
          for (int i = 0; i < n; i++) {
              graph[i] = new ArrayList<>();
          }
          for (int i = 1; i <= m; i++) {
              int s = in.nextInt() - 1, e = in.nextInt() - 1;
              graph[s].add(e);
  //            graph[e].add(s);
          }
          return graph;
      }
  
      private int ni() {
          return in.nextInt();
      }
  
      private long nl() {
          return in.nextLong();
      }
  
      private String ns() {
          return in.next();
      }
  
      private long[] na(int n) {
          long[] A = new long[n];
          for (int i = 0; i < n; i++) {
              A[i] = in.nextLong();
          }
          return A;
      }
  
      private int[] nia(int n) {
          int[] A = new int[n];
          for (int i = 0; i < n; i++) {
              A[i] = in.nextInt();
          }
          return A;
      }
  }
  
  class FastReader {
      BufferedReader br;
      StringTokenizer st;
  
      public FastReader() {
          br = new BufferedReader(new InputStreamReader(System.in));
      }
  
  
      public String next() {
          while (st == null || !st.hasMoreElements())//回车，空行情况
          {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      public int nextInt() {
          return Integer.parseInt(next());
      }
  
      public long nextLong() {
          return Long.parseLong(next());
      }
  }