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.
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);
    }

    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());
    }
}