#include <bits/stdc++.h>
using namespace std;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
template <int m>
struct mod {
  long long x;
  mod() : x(0) {}
  mod(long long xx) : x(xx) {
    if (abs(x) >= m) x %= m;
    if (x < 0) x += m;
  }
  mod operator+(const mod &a) const {
    mod n;
    n.x = x + a.x;
    if (n.x >= m) n.x -= m;
    return n;
  }
  mod operator-(const mod &a) const {
    mod n;
    n.x = x - a.x;
    if (n.x < 0) n.x += m;
    return n;
  }
  mod operator*(const mod &a) const { return x * a.x; }
  mod operator+=(const mod &a) {
    x += a.x;
    if (x >= m) x -= m;
    return *this;
  }
  mod operator-=(const mod &a) {
    x -= a.x;
    if (x < 0) x += m;
    return *this;
  }
  mod operator*=(const mod &a) {
    x = (x * a.x) % m;
    return *this;
  }
  mod pow(long long b) const {
    mod ans = 1;
    mod a = *this;
    while (b > 0) {
      if (b & 1) ans *= a;
      a *= a;
      b /= 2;
    }
    return ans;
  }
  mod inv() const { return pow(m - 2); }
  mod operator/(const mod &a) const { return (*this) * a.inv(); }
  mod operator/=(const mod &a) { return (*this) *= a.inv(); }
  bool operator==(const mod &o) const { return x == o.x; }
  bool operator!=(const mod &o) const { return x != o.x; }
  long long operator()() const { return x; }
};
template <int m>
istream &operator>>(istream &is, mod<m> &n) {
  long long x;
  is >> x;
  n = x;
  return is;
}
template <int m>
ostream &operator<<(ostream &os, const mod<m> &n) {
  return os << n();
}
const int M = 998244353;
using base = mod<M>;
struct pt {
  long long pos;
  int i;
  bool x;
  bool operator<(const pt &o) const { return pos < o.pos; }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> x(n), y(n), s(n);
  vector<pt> ve;
  for (int i = (0); i < (n); i++) {
    cin >> x[i] >> y[i] >> s[i];
    ve.push_back({x[i], i, true});
    ve.push_back({y[i], i, false});
  }
  vector<base> times(n);
  sort(ve.rbegin(), ve.rend());
  base r = 1;
  base ans = 0;
  for (int i = (0); i < (((int)(ve).size())); i++) {
    pt &p = ve[i];
    if (i != 0) ans += r * (ve[i - 1].pos - ve[i].pos);
    if (p.x) {
      if (s[p.i]) {
        times[p.i] = r;
      } else {
        times[p.i] = r - 1;
      }
      r += times[p.i];
    } else {
      r -= times[p.i];
    }
  }
  ans += ve.back().pos + 1;
  cout << ans << '\n';
}

### 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>
using minpq = priority_queue<T, vector<T>, greater<T>>;
template <int m>
struct mod {
  long long x;
  mod() : x(0) {}
  mod(long long xx) : x(xx) {
    if (abs(x) >= m) x %= m;
    if (x < 0) x += m;
  }
  mod operator+(const mod &a) const {
    mod n;
    n.x = x + a.x;
    if (n.x >= m) n.x -= m;
    return n;
  }
  mod operator-(const mod &a) const {
    mod n;
    n.x = x - a.x;
    if (n.x < 0) n.x += m;
    return n;
  }
  mod operator*(const mod &a) const { return x * a.x; }
  mod operator+=(const mod &a) {
    x += a.x;
    if (x >= m) x -= m;
    return *this;
  }
  mod operator-=(const mod &a) {
    x -= a.x;
    if (x < 0) x += m;
    return *this;
  }
  mod operator*=(const mod &a) {
    x = (x * a.x) % m;
    return *this;
  }
  mod pow(long long b) const {
    mod ans = 1;
    mod a = *this;
    while (b > 0) {
      if (b & 1) ans *= a;
      a *= a;
      b /= 2;
    }
    return ans;
  }
  mod inv() const { return pow(m - 2); }
  mod operator/(const mod &a) const { return (*this) * a.inv(); }
  mod operator/=(const mod &a) { return (*this) *= a.inv(); }
  bool operator==(const mod &o) const { return x == o.x; }
  bool operator!=(const mod &o) const { return x != o.x; }
  long long operator()() const { return x; }
};
template <int m>
istream &operator>>(istream &is, mod<m> &n) {
  long long x;
  is >> x;
  n = x;
  return is;
}
template <int m>
ostream &operator<<(ostream &os, const mod<m> &n) {
  return os << n();
}
const int M = 998244353;
using base = mod<M>;
struct pt {
  long long pos;
  int i;
  bool x;
  bool operator<(const pt &o) const { return pos < o.pos; }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> x(n), y(n), s(n);
  vector<pt> ve;
  for (int i = (0); i < (n); i++) {
    cin >> x[i] >> y[i] >> s[i];
    ve.push_back({x[i], i, true});
    ve.push_back({y[i], i, false});
  }
  vector<base> times(n);
  sort(ve.rbegin(), ve.rend());
  base r = 1;
  base ans = 0;
  for (int i = (0); i < (((int)(ve).size())); i++) {
    pt &p = ve[i];
    if (i != 0) ans += r * (ve[i - 1].pos - ve[i].pos);
    if (p.x) {
      if (s[p.i]) {
        times[p.i] = r;
      } else {
        times[p.i] = r - 1;
      }
      r += times[p.i];
    } else {
      r -= times[p.i];
    }
  }
  ans += ve.back().pos + 1;
  cout << ans << '\n';
}

```