#include <bits/stdc++.h>
using namespace std;
template <const int32_t MOD>
struct ModInt {
  int32_t x;
  ModInt() : x(0) {}
  ModInt(long long u) : x(u % MOD) {
    x %= MOD;
    if (x < 0) x += MOD;
  }
  friend bool operator==(const ModInt &a, const ModInt &b) {
    return a.x == b.x;
  }
  friend bool operator!=(const ModInt &a, const ModInt &b) {
    return a.x != b.x;
  }
  friend bool operator<(const ModInt &a, const ModInt &b) { return a.x < b.x; }
  friend bool operator>(const ModInt &a, const ModInt &b) { return a.x > b.x; }
  friend bool operator<=(const ModInt &a, const ModInt &b) {
    return a.x <= b.x;
  }
  friend bool operator>=(const ModInt &a, const ModInt &b) {
    return a.x >= b.x;
  }
  static ModInt sign(long long k) {
    return ((k & 1) ? ModInt(MOD - 1) : ModInt(1));
  }
  ModInt &operator+=(const ModInt &m) {
    x += m.x;
    if (x >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator-=(const ModInt &m) {
    x -= m.x;
    if (x < 0LL) x += MOD;
    return *this;
  }
  ModInt &operator*=(const ModInt &m) {
    x = (1LL * x * m.x) % MOD;
    return *this;
  }
  friend ModInt operator-(const ModInt &a) {
    ModInt res(a);
    if (res.x) res.x = MOD - res.x;
    return res;
  }
  friend ModInt operator+(const ModInt &a, const ModInt &b) {
    return ModInt(a) += ModInt(b);
  }
  friend ModInt operator-(const ModInt &a, const ModInt &b) {
    return ModInt(a) -= ModInt(b);
  }
  friend ModInt operator*(const ModInt &a, const ModInt &b) {
    return ModInt(a) *= ModInt(b);
  }
  static long long fp(long long u, long long k) {
    long long res = 1LL;
    while (k > 0LL) {
      if (k & 1LL) res = (res * u) % MOD;
      u = (u * u) % MOD;
      k /= 2LL;
    }
    return res;
  }
  static constexpr long long mod() { return MOD; }
  ModInt fastpow(long long k) { return ModInt(fp(x, k)); }
  ModInt inv() {
    assert(x);
    return ModInt(fp(x, MOD - 2));
  }
  ModInt &operator/=(const ModInt &m) { return *this *= ModInt(m).inv(); }
  friend ModInt operator/(const ModInt &a, const ModInt &b) {
    return ModInt(a) *= ModInt(b).inv();
  }
  friend ostream &operator<<(ostream &out, const ModInt &a) {
    return out << a.x;
  }
  friend istream &operator>>(istream &in, ModInt &a) { return in >> a.x; }
};
const long long MOD = 998244353;
using Mint = ModInt<MOD>;
template <class T>
class Fenwick {
 public:
  long long lim;
  vector<T> bit;
  Fenwick(long long n) : lim(n + 1), bit(lim) {}
  void upd(long long pos, T val) {
    for (pos++; pos < lim; pos += pos & -pos) bit[pos] += val;
  }
  T sum(long long r) {
    T ret = 0;
    for (; r; r -= r & -r) ret += bit[r];
    return ret;
  }
  T sum(long long l, long long r) { return sum(r) - sum(l); }
};
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long nbPortails;
  cin >> nbPortails;
  vector<long long> depart(nbPortails), arrive(nbPortails), state(nbPortails);
  vector<Mint> cost(nbPortails);
  for (long long i = 0; i < nbPortails; ++i)
    cin >> depart[i] >> arrive[i] >> state[i];
  Fenwick<Mint> fen(nbPortails);
  for (long long iPortail = 0; iPortail < nbPortails; ++iPortail) {
    long long fst =
        lower_bound(depart.begin(), depart.end(), arrive[iPortail]) -
        depart.begin();
    cost[iPortail] =
        depart[iPortail] - arrive[iPortail] + fen.sum(fst, iPortail);
    fen.upd(iPortail, cost[iPortail]);
  }
  Mint ret = depart.back() + 1;
  for (long long i = 0; i < nbPortails; ++i)
    if (state[i]) ret += cost[i];
  cout << ret << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <const int32_t MOD>
struct ModInt {
  int32_t x;
  ModInt() : x(0) {}
  ModInt(long long u) : x(u % MOD) {
    x %= MOD;
    if (x < 0) x += MOD;
  }
  friend bool operator==(const ModInt &a, const ModInt &b) {
    return a.x == b.x;
  }
  friend bool operator!=(const ModInt &a, const ModInt &b) {
    return a.x != b.x;
  }
  friend bool operator<(const ModInt &a, const ModInt &b) { return a.x < b.x; }
  friend bool operator>(const ModInt &a, const ModInt &b) { return a.x > b.x; }
  friend bool operator<=(const ModInt &a, const ModInt &b) {
    return a.x <= b.x;
  }
  friend bool operator>=(const ModInt &a, const ModInt &b) {
    return a.x >= b.x;
  }
  static ModInt sign(long long k) {
    return ((k & 1) ? ModInt(MOD - 1) : ModInt(1));
  }
  ModInt &operator+=(const ModInt &m) {
    x += m.x;
    if (x >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator-=(const ModInt &m) {
    x -= m.x;
    if (x < 0LL) x += MOD;
    return *this;
  }
  ModInt &operator*=(const ModInt &m) {
    x = (1LL * x * m.x) % MOD;
    return *this;
  }
  friend ModInt operator-(const ModInt &a) {
    ModInt res(a);
    if (res.x) res.x = MOD - res.x;
    return res;
  }
  friend ModInt operator+(const ModInt &a, const ModInt &b) {
    return ModInt(a) += ModInt(b);
  }
  friend ModInt operator-(const ModInt &a, const ModInt &b) {
    return ModInt(a) -= ModInt(b);
  }
  friend ModInt operator*(const ModInt &a, const ModInt &b) {
    return ModInt(a) *= ModInt(b);
  }
  static long long fp(long long u, long long k) {
    long long res = 1LL;
    while (k > 0LL) {
      if (k & 1LL) res = (res * u) % MOD;
      u = (u * u) % MOD;
      k /= 2LL;
    }
    return res;
  }
  static constexpr long long mod() { return MOD; }
  ModInt fastpow(long long k) { return ModInt(fp(x, k)); }
  ModInt inv() {
    assert(x);
    return ModInt(fp(x, MOD - 2));
  }
  ModInt &operator/=(const ModInt &m) { return *this *= ModInt(m).inv(); }
  friend ModInt operator/(const ModInt &a, const ModInt &b) {
    return ModInt(a) *= ModInt(b).inv();
  }
  friend ostream &operator<<(ostream &out, const ModInt &a) {
    return out << a.x;
  }
  friend istream &operator>>(istream &in, ModInt &a) { return in >> a.x; }
};
const long long MOD = 998244353;
using Mint = ModInt<MOD>;
template <class T>
class Fenwick {
 public:
  long long lim;
  vector<T> bit;
  Fenwick(long long n) : lim(n + 1), bit(lim) {}
  void upd(long long pos, T val) {
    for (pos++; pos < lim; pos += pos & -pos) bit[pos] += val;
  }
  T sum(long long r) {
    T ret = 0;
    for (; r; r -= r & -r) ret += bit[r];
    return ret;
  }
  T sum(long long l, long long r) { return sum(r) - sum(l); }
};
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long nbPortails;
  cin >> nbPortails;
  vector<long long> depart(nbPortails), arrive(nbPortails), state(nbPortails);
  vector<Mint> cost(nbPortails);
  for (long long i = 0; i < nbPortails; ++i)
    cin >> depart[i] >> arrive[i] >> state[i];
  Fenwick<Mint> fen(nbPortails);
  for (long long iPortail = 0; iPortail < nbPortails; ++iPortail) {
    long long fst =
        lower_bound(depart.begin(), depart.end(), arrive[iPortail]) -
        depart.begin();
    cost[iPortail] =
        depart[iPortail] - arrive[iPortail] + fen.sum(fst, iPortail);
    fen.upd(iPortail, cost[iPortail]);
  }
  Mint ret = depart.back() + 1;
  for (long long i = 0; i < nbPortails; ++i)
    if (state[i]) ret += cost[i];
  cout << ret << endl;
}

```