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.*;
import java.text.*;

public class CF_1552_F{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int N = ni();
        int[][] seg = new int[N][];
        int[] list = new int[N*2];
        for(int i = 0; i< N; i++){
            int s = ni(), t = ni();
            seg[i] = new int[]{s, t, s-t, ni()};
            list[i<<1] = s;
            list[i<<1|1] = t;
        }
        Arrays.sort(list);
        for(int i = 0; i< N; i++)for(int t = 0; t< 2; t++)seg[i][t] = Arrays.binarySearch(list, seg[i][t]);
        long MOD = 998244353;
        long[] f = new long[2*N];
        FenwickTree ft = new FenwickTree(2*N);
        for(int i = 0; i< N; i++){
            f[i] = (seg[i][2] + ft.sum(seg[i][1], seg[i][0]))%MOD;
            ft.add(seg[i][0], f[i]);
        }
        long ans = (list[2*N-1]+1)%MOD;
        for(int i = 0; i< N; i++){
            if(seg[i][3] == 1){
                ans += f[i];
                if(ans >= MOD)ans -= MOD;
            }
        }
        pn(ans);
    }
    class FenwickTree {
        long[] bit;
        int N;
        //uses 0-based indexing
        public FenwickTree(int N){
            this.N = N;
            bit = new long[N];
        }
        void add(int idx, long delta){
            for (; idx < N; idx = idx | (idx + 1))
                bit[idx] += delta;
        }
        long sum(int r){
            if(r >= N)r = N-1;
            long ret = 0;
            for (; r >= 0; r = (r & (r + 1)) - 1)
                ret += bit[r];
            return ret;
        }
        long sum(int l, int r){return sum(r)-sum(l-1);}
    }

    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
    final long IINF = (long)1e17;
    final int INF = (int)1e9+2;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
    static boolean multipleTC = false, memory = true, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        long ct = System.currentTimeMillis();
        if (fileIO) {
            in = new FastReader("");
            out = new PrintWriter("");
        } else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = multipleTC? ni():1;
        pre();
        for (int t = 1; t <= T; t++) solve(t);
        out.flush();
        out.close();
        System.err.println(System.currentTimeMillis() - ct);
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1552_F().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
        else new CF_1552_F().run();
    }
    int[][] make(int n, int e, int[] from, int[] to, boolean f){
        int[][] g = new int[n][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            if(f)g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
        int[][][] g = new int[n][][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
            if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
        }
        return g;
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object... o){for(Object oo:o)out.print(oo+" ");}
    void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
    void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str;
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}