#include <bits/stdc++.h>
using namespace std;
long long int min(long long int a, long long int b) { return (a > b) ? b : a; }
long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
struct node {
  long long int x;
  node() { x = 0; }
  node operator+(const node& other) {
    node sum = node();
    sum.x = (x + other.x) % 998244353;
    return sum;
  }
};
struct SegTree {
  vector<node> t;
  long long int n;
  SegTree(long long int n) {
    this->n = n;
    t.assign(4 * n, node());
  }
  SegTree(vector<long long int>& a) : SegTree(a.size()) { build(a, 1, 1, n); }
  void build(vector<long long int>& a, int v, int tl, int tr) {
    if (tl == tr)
      t[v].x = a[tl - 1];
    else {
      int tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = t[v * 2] + t[v * 2 + 1];
    }
  }
  node sum(int v, int tl, int tr, int l, int r) {
    if (l > r) return node();
    if (l == tl && r == tr) return t[v];
    int tm = (tl + tr) / 2;
    return sum(v * 2, tl, tm, l, min(r, tm)) +
           sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  }
  long long int sum(int l, int r) { return sum(1, 1, n, l, r).x; }
  void update(int v, int tl, int tr, int pos, long long int new_val) {
    if (tl == tr)
      t[v].x = new_val;
    else {
      int tm = (tl + tr) / 2;
      if (pos <= tm)
        update(2 * v, tl, tm, pos, new_val);
      else
        update(2 * v + 1, tm + 1, tr, pos, new_val);
      t[v] = t[2 * v] + t[2 * v + 1];
    }
  }
  void update(int idx, long long int a) { update(1, 1, n, idx, a); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int te = 1;
  while (te--) {
    long long int n;
    cin >> n;
    long long int x[n], y[n], f[n];
    for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> f[i];
    long long int ans = (x[n - 1] + 1) % 998244353;
    map<long long int, long long int> m;
    for (int i = 0; i < n; i++) m[x[i]]++, m[y[i]]++;
    int c = 0;
    for (auto i : m) m[i.first] = ++c;
    SegTree ST(c + 5);
    for (int i = 0; i < n; i++) {
      long long int s = ST.sum(m[y[i]], m[x[i]]);
      (s += x[i] - y[i]) %= 998244353;
      ST.update(m[x[i]], s);
      if (f[i]) (ans += s) %= 998244353;
    }
    cout << ans << '\n';
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int min(long long int a, long long int b) { return (a > b) ? b : a; }
long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
struct node {
  long long int x;
  node() { x = 0; }
  node operator+(const node& other) {
    node sum = node();
    sum.x = (x + other.x) % 998244353;
    return sum;
  }
};
struct SegTree {
  vector<node> t;
  long long int n;
  SegTree(long long int n) {
    this->n = n;
    t.assign(4 * n, node());
  }
  SegTree(vector<long long int>& a) : SegTree(a.size()) { build(a, 1, 1, n); }
  void build(vector<long long int>& a, int v, int tl, int tr) {
    if (tl == tr)
      t[v].x = a[tl - 1];
    else {
      int tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = t[v * 2] + t[v * 2 + 1];
    }
  }
  node sum(int v, int tl, int tr, int l, int r) {
    if (l > r) return node();
    if (l == tl && r == tr) return t[v];
    int tm = (tl + tr) / 2;
    return sum(v * 2, tl, tm, l, min(r, tm)) +
           sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  }
  long long int sum(int l, int r) { return sum(1, 1, n, l, r).x; }
  void update(int v, int tl, int tr, int pos, long long int new_val) {
    if (tl == tr)
      t[v].x = new_val;
    else {
      int tm = (tl + tr) / 2;
      if (pos <= tm)
        update(2 * v, tl, tm, pos, new_val);
      else
        update(2 * v + 1, tm + 1, tr, pos, new_val);
      t[v] = t[2 * v] + t[2 * v + 1];
    }
  }
  void update(int idx, long long int a) { update(1, 1, n, idx, a); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int te = 1;
  while (te--) {
    long long int n;
    cin >> n;
    long long int x[n], y[n], f[n];
    for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> f[i];
    long long int ans = (x[n - 1] + 1) % 998244353;
    map<long long int, long long int> m;
    for (int i = 0; i < n; i++) m[x[i]]++, m[y[i]]++;
    int c = 0;
    for (auto i : m) m[i.first] = ++c;
    SegTree ST(c + 5);
    for (int i = 0; i < n; i++) {
      long long int s = ST.sum(m[y[i]], m[x[i]]);
      (s += x[i] - y[i]) %= 998244353;
      ST.update(m[x[i]], s);
      if (f[i]) (ans += s) %= 998244353;
    }
    cout << ans << '\n';
  }
}

```