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.util.Map.Entry;

import java.math.*;
import java.io.*;

public class Main {
	public static void main(String[] args) throws FileNotFoundException {
		InputReader in = new InputReader(System.in);
		// Scanner in = new Scanner(System.in);
		// Scanner in = new Scanner(new BufferedReader(new
		// InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(System.out);
		// InputReader in = new InputReader(new
		// File("ethan_traverses_a_tree.txt"));
		// PrintWriter out = new PrintWriter(new
		// File("ethan_traverses_a_tree-output.txt"));

		final int mod = 998244353;

		int n = in.nextInt();

		int[] x = new int[n];
		int[] y = new int[n];
		int[] s = new int[n];

		for (int i = 0; i < n; i++) {
			x[i] = in.nextInt();
			y[i] = in.nextInt();
			s[i] = in.nextInt();
		}

		TreeMap<Integer, Integer> treeMap = new TreeMap<Integer, Integer>();
		long[] dp = new long[n];
		long[] sum = new long[n];
		for (int i = 0; i < n; i++) {
			if (treeMap.ceilingKey(y[i]) == null) {
				dp[i] = (x[i] - y[i]) % mod;
			} else {
				int firstx = treeMap.ceilingKey(y[i]);
				int pos = treeMap.get(firstx);

				if (pos == 0) {
					dp[i] = (sum[i - 1] + x[i] - y[i]) % mod;
				} else {
					dp[i] = (sum[i - 1] - sum[pos - 1] + x[i] - y[i] + mod) % mod;
				}
			}

			if (i == 0) {
				sum[i] = dp[i];
			} else {
				sum[i] = (sum[i - 1] + dp[i]) % mod;
			}
			treeMap.put(x[i], i);
		}

		long ans = (x[n - 1] + 1) % mod;
		for (int i = 0; i < n; i++) {
			if (s[i] == 1) {
				ans = (ans + dp[i]) % mod;
			}
		}

		out.printf("%d\n", ans);

		out.close();
	}

	static class InputReader {
		BufferedReader br;
		StringTokenizer st;

		public InputReader(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		public boolean hasNext() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}