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.io.*;
  import java.util.*;
  public class Main {
  	static int bs(int start,int end,int[][] l,int val) {
  		if(start>end) return -1;
  		int mid=(start+end)/2;
  		if(l[mid][0]<val) {
  			return Math.max(mid, bs(mid+1,end,l,val));
  		}
  		return bs(start,mid-1,l,val);
  	}
  	public static void main(String[] args) throws IOException 
  	{ 
  		FastScanner f = new FastScanner(); 
  		int ttt=1;
  //		ttt=f.nextInt();
  		PrintWriter out=new PrintWriter(System.out);
  		outer:for(int tt=0;tt<ttt;tt++) {
  			int n=f.nextInt(); 
  			int[][] l=new int[n][3];
  			for(int i=0;i<n;i++) {
  				l[i]=f.readArray(3);
  			}
  			long[] steps=new long[n+1];
  			long mod=998244353;
  			long[] cum=new long[n+1];
  			for(int i=1;i<=n;i++) {
  				long num=(l[i-1][0]-l[i-1][1])%mod;
  				int ind=bs(0,n-1,l,l[i-1][1]);
  				num=(num+(cum[i-1]-cum[ind+1])%mod)%mod;
  				steps[i]=num;
  				cum[i]=(cum[i-1]+steps[i])%mod;
  			}
  			long ans=0;
  //			System.out.println(Arrays.toString(steps));
  //			System.out.println(Arrays.toString(cum));
  			int prev=0;
  			for(int i=0;i<n;i++) {
  				ans=(ans+(mod+l[i][0]-prev)%mod)%mod;
  				prev=l[i][0];
  				if(l[i][2]==1) {
  					ans=(ans+steps[i+1])%mod;
  				}
  			}
  			System.out.println((mod+ans+1)%mod);
  		}
  		out.close();
  	} 
  	static void sort(int[] p) {
          ArrayList<Integer> q = new ArrayList<>();
          for (int i: p) q.add( i);
          Collections.sort(q);
          for (int i = 0; i < p.length; i++) p[i] = q.get(i);
      }
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		long[] readLongArray(int n) {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  	}
  }