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.
#include "bits/stdc++.h"
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <climits>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <utility>
#include <complex>
#include <chrono>
#include <stdlib.h>
using namespace std;
using namespace std::chrono;
typedef long long int lli;
typedef long double ld;
typedef vector<lli> vl;
typedef vector<vl> vvl;
typedef vector<ld> vd;
typedef map<lli,lli> ml;
typedef unordered_map<lli,lli> uml;
typedef pair<lli,lli> pl;
typedef pair<pl,lli> ppl;
typedef set<lli> sl;
typedef multiset<lli> msl;
typedef vector<pl> vp;
typedef queue<lli> ql;
typedef priority_queue<lli> pql;
typedef list<lli> ll;
typedef vector<ll> graph;
# define pb push_back
# define mp make_pair
# define f first
# define s second
# define itr iterator
# define MOD1 1000000007
# define MOD 998244353
# define INF 1000000000000000000
# define all(x) x.begin(), x.end()
# define rall(a) a.begin(), a.end(), greater<lli>()
# define ins insert
# define ms(a,x) memset(a, x, sizeof(a))
# define rep(i,n) for(i=0;i<n;i++)
# define wrep(i,a,b) for(i=a;i<b;i++)
# define M 300005
const string alpha = "abcdefghijklmnopqrstuvwxyz";

void solve(){
    lli n, i, m=0, q=0, mx=-INF, mn=INF, k=0, j=0, c=0, d=0, t=0, sum=0, pdt=1, l=0, r=0;
    cin >> n;

    lli at[n], to[n], cond[n];
    ml track; sl ports;
    for(i=0;i<n;i++){
      cin >> l >> r >> c;
      at[i] = l; to[i] = r; cond[i] = c;
      track[l] = i;
      ports.insert(l);
    }
    lli dp[n] = {0}, pre[n] = {0};
    for(i=0;i<n;i++){
      l = at[i]; r = to[i];
      auto j = ports.lower_bound(r);
      dp[i] += *j-r; dp[i] %= MOD;
      k = track[*j];
      if(i==k){
        if(i!=0) pre[i] = pre[i-1];
        pre[i] += dp[i]; pre[i] %= MOD;
        continue;
      }
      if(i!=0) dp[i] += pre[i-1];
      if(k!=0) dp[i] -= pre[k-1];
      if(i!=0) dp[i] += at[i]-at[k];
      if(i!=0) pre[i] = pre[i-1];
      pre[i] += dp[i];
      dp[i] %= MOD; pre[i] %= MOD;
    }
    l = 0;
    for(i=0;i<n;i++){
      sum += at[i]-l;
      l = at[i];
      if(cond[i]==1) sum += dp[i];
      sum %= MOD;
    }
    cout<<(sum+1)%MOD<<endl;
    return;
}



int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    lli t=1;
    while(t--){
        solve();
    }
    return 0;
}
