1552_F. Telepanting_75
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>
using namespace std;
int n;
const int mod = 998244353;
struct wocao {
  int x, y, s;
} a[200020];
int aa[400040];
map<int, int> mp;
int bb[400040];
int c[400040];
int lowbit(int x) { return x & (-x); }
int query(int x) {
  int ret = 0;
  while (x) (ret += c[x]) %= mod, x -= lowbit(x);
  return ret;
}
void add(int x, int y) {
  while (x <= n * 2) (c[x] += y) %= mod, x += lowbit(x);
}
int t[400040];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].s), aa[i * 2 - 1] = a[i].x,
                                                           aa[i * 2] = a[i].y;
  sort(aa + 1, aa + n * 2 + 1);
  int tttt = 0;
  for (int i = 1; i <= 2 * n; i++)
    if (aa[i] != aa[i - 1]) mp[aa[i]] = ++tttt, bb[tttt] = aa[i];
  for (int i = 1; i <= n; i++) a[i].x = mp[a[i].x], a[i].y = mp[a[i].y];
  int temp;
  for (int i = n; i >= 1; i--) {
    if (a[i].s == 1) {
      t[i] = (query(a[i].x) + 1) % mod;
    } else {
      t[i] = query(a[i].x);
    }
    add(a[i].y, t[i]);
  }
  int ans = (bb[a[n].x] + 1) % mod;
  for (int i = 1; i <= n; i++) {
    long long l = bb[a[i].x] - bb[a[i].y];
    (l *= t[i]) %= mod;
    (ans += l) %= mod;
  }
  cout << ans;
  return 0;
}
