#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e16;
const int mod = 998244353;
void solve() {
  int n;
  cin >> n;
  vector<int> x(n);
  vector<int> y(n);
  vector<int> s(n);
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> s[i];
  vector<long long> ans(n, 0);
  vector<long long> f(n, 0);
  ans[0] = x[0] % mod;
  f[0] = (x[0] - y[0]) % mod;
  if (s[0] == 1) ans[0] = (ans[0] + f[0]) % mod;
  vector<long long> pref(n + 1, 0);
  pref[1] = f[0];
  for (int i = 1; i < n; i++) {
    int l = 0, r = i - 1;
    while (l < r) {
      int mid = (l + r) / 2;
      if (x[mid] < y[i])
        l = mid + 1;
      else
        r = mid;
    }
    if (x[l] > y[i]) f[i] = (pref[i] - pref[l] + mod) % mod;
    f[i] = (f[i] + x[i] - y[i]) % mod;
    pref[i + 1] = (pref[i] + f[i]) % mod;
    ans[i] = (ans[i - 1] + x[i] - x[i - 1]) % mod;
    if (s[i] == 1) ans[i] = (ans[i] + f[i]) % mod;
  }
  cout << (ans[n - 1] + 1) % mod << "\n";
  return;
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test_cnt = 1;
  for (int test = 1; test <= test_cnt; test++) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
const long long inf = 1e16;
const int mod = 998244353;
void solve() {
  int n;
  cin >> n;
  vector<int> x(n);
  vector<int> y(n);
  vector<int> s(n);
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> s[i];
  vector<long long> ans(n, 0);
  vector<long long> f(n, 0);
  ans[0] = x[0] % mod;
  f[0] = (x[0] - y[0]) % mod;
  if (s[0] == 1) ans[0] = (ans[0] + f[0]) % mod;
  vector<long long> pref(n + 1, 0);
  pref[1] = f[0];
  for (int i = 1; i < n; i++) {
    int l = 0, r = i - 1;
    while (l < r) {
      int mid = (l + r) / 2;
      if (x[mid] < y[i])
        l = mid + 1;
      else
        r = mid;
    }
    if (x[l] > y[i]) f[i] = (pref[i] - pref[l] + mod) % mod;
    f[i] = (f[i] + x[i] - y[i]) % mod;
    pref[i + 1] = (pref[i] + f[i]) % mod;
    ans[i] = (ans[i - 1] + x[i] - x[i - 1]) % mod;
    if (s[i] == 1) ans[i] = (ans[i] + f[i]) % mod;
  }
  cout << (ans[n - 1] + 1) % mod << "\n";
  return;
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test_cnt = 1;
  for (int test = 1; test <= test_cnt; test++) {
    solve();
  }
  return 0;
}

```