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>
  #pragma GCC optimize("-O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  mt19937_64 rnd(time(0));
  long long MOD = 998244353;
  const long double E = 1e-10;
  long long n, m, k, T, pr[500001];
  long long binpow(long long h, long long r, long long md) {
    long long l = 1;
    while (r) {
      if (r & 1) l = (l * h) % md;
      h = (h * h) % md;
      r /= 2;
    }
    return l;
  }
  long long read() {
    char c = getchar();
    while (!isdigit(c)) {
      c = getchar();
    }
    long long res = 0;
    while (isdigit(c)) {
      res = res * 10 + c - '0';
      c = getchar();
    }
    return res;
  }
  long long gcd(long long x, long long y) {
    while (x && y) {
      if (x < y) swap(x, y);
      x %= y;
    }
    return x + y;
  }
  long long d[2000001];
  void push(int h) {
    if (d[h]) {
      d[h * 2] = (d[h * 2] + d[h]) % MOD;
      d[h * 2 + 1] = (d[h * 2 + 1] + d[h]) % MOD;
      d[h] = 0;
    }
  }
  void update(int h, int l, int r, int x, int y, int k) {
    if (x > y) return;
    if (l == x && y == r) {
      d[h] = (d[h] + k) % MOD;
      return;
    }
    push(h);
    int w = (l + r) / 2;
    update(h * 2, l, w, x, min(y, w), k);
    update(h * 2 + 1, w + 1, r, max(x, w + 1), y, k);
  }
  long long get(int h, int l, int r, int x) {
    if (l == r) {
      return d[h];
    }
    push(h);
    int w = (l + r) / 2;
    if (x <= w)
      return get(h * 2, l, w, x);
    else
      return get(h * 2 + 1, w + 1, r, x);
  }
  void solve() {
    cin >> n;
    vector<long long> v;
    vector<long long> zz, qq;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      long long x, y, z;
      cin >> x >> y >> z;
      ans = x + 1;
      v.push_back(x);
      zz.push_back(z);
      qq.push_back((x - y) % MOD);
      long long o = lower_bound(v.begin(), v.end(), y) - v.begin();
      pr[i] = o;
    }
    for (int i = n - 1; i > -1; i--) {
      if (zz[i] == 1) update(1, 0, n - 1, i, i, 1);
      long long o = get(1, 0, n - 1, i);
      ans = (ans + qq[i] * o) % MOD;
      update(1, 0, n - 1, pr[i], i - 1, o);
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  