#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
const int P = 998244353;
int n, X[200005], Y[200005];
long long dp[200005], sdp[200005];
char v[200005];
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(X[i]), read(Y[i]), read(v[i]);
  sdp[0] = dp[0] = (X[1] << 1) - Y[1];
  for (int i = 1; i < n; i++) {
    dp[i] = (X[i + 1] - X[i]) << 1;
    int wh = lower_bound(X + 1, X + n + 1, Y[i + 1]) - X - 1;
    dp[i] = (dp[i] + sdp[i - 1] - sdp[wh - 1] + X[wh] - Y[i + 1]) % P,
    sdp[i] = (sdp[i - 1] + dp[i]) % P;
  }
  long long rs = 0;
  for (int i = 1; i <= n; i++)
    if (!v[i])
      rs = (rs + X[i] - X[i - 1]) % P;
    else
      rs = (rs + dp[i - 1]) % P;
  return printf("%lld\n", (rs + P + 1) % P), 0;
}

### Prompt

Create a solution in CPP for 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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
const int P = 998244353;
int n, X[200005], Y[200005];
long long dp[200005], sdp[200005];
char v[200005];
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(X[i]), read(Y[i]), read(v[i]);
  sdp[0] = dp[0] = (X[1] << 1) - Y[1];
  for (int i = 1; i < n; i++) {
    dp[i] = (X[i + 1] - X[i]) << 1;
    int wh = lower_bound(X + 1, X + n + 1, Y[i + 1]) - X - 1;
    dp[i] = (dp[i] + sdp[i - 1] - sdp[wh - 1] + X[wh] - Y[i + 1]) % P,
    sdp[i] = (sdp[i - 1] + dp[i]) % P;
  }
  long long rs = 0;
  for (int i = 1; i <= n; i++)
    if (!v[i])
      rs = (rs + X[i] - X[i - 1]) % P;
    else
      rs = (rs + dp[i - 1]) % P;
  return printf("%lld\n", (rs + P + 1) % P), 0;
}

```