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: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  using matr = vector<vector<ll>>;
  using dbl = long double;
  using ui = unsigned int;
  using ld = long double;
  mt19937 rnd;
  ll random(ll limit) {
    uniform_int_distribution<ll> uniform_dist(0, limit);
    return uniform_dist(rnd);
  }
  ll mod = 998244353;
  ll bpow(ll a, ll p) {
    ll res = 1;
    while (p > 0) {
      if (p % 2) res *= a;
      p /= 2;
      a *= a;
      res %= mod;
      a %= mod;
    }
    return res;
  }
  ll inv(ll a) { return bpow(a, mod - 2); }
  ll gcd(ll a, ll b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  void solve() {
    int n;
    cin >> n;
    vector<ll> x(n), y(n), s(n);
    vector<vector<ll>> kek;
    for (int i = 0; i < n; i++) {
      cin >> x[i] >> y[i] >> s[i];
      kek.push_back({y[i], 0, i});
      kek.push_back({x[i], 1, i});
    }
    sort(kek.begin(), kek.end());
    map<ll, ll> pref;
    ll cur = 0;
    ll ans = 0;
    ll last = 0;
    for (int i = 0; i < kek.size(); i++) {
      if (kek[i][1] == 0) {
        cur += kek[i][0] - last;
        cur %= mod;
        cur += mod;
        cur %= mod;
        pref[kek[i][0]] = cur;
      } else {
        cur += kek[i][0] - last;
        cur %= mod;
        cur += mod;
        cur %= mod;
        ll beg = y[kek[i][2]];
        ll len = (cur - pref[beg] + mod) % mod;
        if (s[kek[i][2]] == 0) {
          ans += mod - len;
          ans %= mod;
        }
        pref[kek[i][0]] = (cur + len) % mod;
        cur += len;
        cur %= mod;
      }
      last = kek[i][0];
      last %= mod;
    }
    ll ans1 = pref[kek[kek.size() - 1][0]] + 1;
    ans1 %= mod;
    ans += ans1;
    ans %= mod;
    cout << ans << '\n';
  }
  int main() {
    int t;
    t = 1;
    for (int i = 0; i < t; i++) {
      solve();
    }
  }
  