#include <bits/stdc++.h>
using namespace std;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  friend modnum inv(const modnum& m) { return m.inv(); }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  friend modnum neg(const modnum& m) { return m.neg(); }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
template <typename T>
T pow(T a, long long b) {
  assert(b >= 0);
  T r = 1;
  while (b) {
    if (b & 1) r *= a;
    b >>= 1;
    a *= a;
  }
  return r;
}
using num = modnum<998244353>;
vector<num> fact, ifact;
void init() {
  int N = 1100000;
  fact = {1};
  for (int i = 1; i < N; i++) fact.push_back(i * fact[i - 1]);
  ifact.resize(N);
  ifact.back() = 1 / fact.back();
  for (int i = N - 1; i > 0; i--) ifact[i - 1] = i * ifact[i];
}
num ncr(int n, int k) {
  if (k < 0 || k > n) return 0;
  return fact[n] * ifact[k] * ifact[n - k];
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  init();
  int n;
  cin >> n;
  vector<int> x(n), y(n), s(n);
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> s[i];
  num ans = x.back() + 1;
  vector<num> subt(n, 0);
  num cur = 1;
  for (int i = n - 1; i >= 0; i--) {
    num go = cur - (1 - s[i]);
    ans += go * (x[i] - y[i]);
    cur += go;
    int st = -1;
    int en = n;
    while (st + 1 < en) {
      int mi = (st + en) / 2;
      if (x[mi] > y[i]) {
        en = mi;
      } else {
        st = mi;
      }
    }
    assert(en <= i);
    subt[en] += go;
    cur -= subt[i];
  }
  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 <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  friend modnum inv(const modnum& m) { return m.inv(); }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  friend modnum neg(const modnum& m) { return m.neg(); }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
template <typename T>
T pow(T a, long long b) {
  assert(b >= 0);
  T r = 1;
  while (b) {
    if (b & 1) r *= a;
    b >>= 1;
    a *= a;
  }
  return r;
}
using num = modnum<998244353>;
vector<num> fact, ifact;
void init() {
  int N = 1100000;
  fact = {1};
  for (int i = 1; i < N; i++) fact.push_back(i * fact[i - 1]);
  ifact.resize(N);
  ifact.back() = 1 / fact.back();
  for (int i = N - 1; i > 0; i--) ifact[i - 1] = i * ifact[i];
}
num ncr(int n, int k) {
  if (k < 0 || k > n) return 0;
  return fact[n] * ifact[k] * ifact[n - k];
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  init();
  int n;
  cin >> n;
  vector<int> x(n), y(n), s(n);
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> s[i];
  num ans = x.back() + 1;
  vector<num> subt(n, 0);
  num cur = 1;
  for (int i = n - 1; i >= 0; i--) {
    num go = cur - (1 - s[i]);
    ans += go * (x[i] - y[i]);
    cur += go;
    int st = -1;
    int en = n;
    while (st + 1 < en) {
      int mi = (st + en) / 2;
      if (x[mi] > y[i]) {
        en = mi;
      } else {
        st = mi;
      }
    }
    assert(en <= i);
    subt[en] += go;
    cur -= subt[i];
  }
  cout << ans << '\n';
}

```