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>
#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[100001];
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() {
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
