#include <bits/stdc++.h>
namespace atcoder {
namespace internal {
int ceil_pow2(int n) {
  int x = 0;
  while ((1U << x) < (unsigned int)(n)) x++;
  return x;
}
int bsf(unsigned int n) { return __builtin_ctz(n); }
}  // namespace internal
}  // namespace atcoder
namespace atcoder {
template <class S, S (*op)(S, S), S (*e)()>
struct segtree {
 public:
  segtree() : segtree(0) {}
  segtree(int n) : segtree(std::vector<S>(n, e())) {}
  segtree(const std::vector<S>& v) : _n(int(v.size())) {
    log = internal::ceil_pow2(_n);
    size = 1 << log;
    d = std::vector<S>(2 * size, e());
    for (int i = 0; i < _n; i++) d[size + i] = v[i];
    for (int i = size - 1; i >= 1; i--) {
      update(i);
    }
  }
  void set(int p, S x) {
    assert(0 <= p && p < _n);
    p += size;
    d[p] = x;
    for (int i = 1; i <= log; i++) update(p >> i);
  }
  S get(int p) {
    assert(0 <= p && p < _n);
    return d[p + size];
  }
  S prod(int l, int r) {
    assert(0 <= l && l <= r && r <= _n);
    S sml = e(), smr = e();
    l += size;
    r += size;
    while (l < r) {
      if (l & 1) sml = op(sml, d[l++]);
      if (r & 1) smr = op(d[--r], smr);
      l >>= 1;
      r >>= 1;
    }
    return op(sml, smr);
  }
  S all_prod() { return d[1]; }
  template <bool (*f)(S)>
  int max_right(int l) {
    return max_right(l, [](S x) { return f(x); });
  }
  template <class F>
  int max_right(int l, F f) {
    assert(0 <= l && l <= _n);
    assert(f(e()));
    if (l == _n) return _n;
    l += size;
    S sm = e();
    do {
      while (l % 2 == 0) l >>= 1;
      if (!f(op(sm, d[l]))) {
        while (l < size) {
          l = (2 * l);
          if (f(op(sm, d[l]))) {
            sm = op(sm, d[l]);
            l++;
          }
        }
        return l - size;
      }
      sm = op(sm, d[l]);
      l++;
    } while ((l & -l) != l);
    return _n;
  }
  template <bool (*f)(S)>
  int min_left(int r) {
    return min_left(r, [](S x) { return f(x); });
  }
  template <class F>
  int min_left(int r, F f) {
    assert(0 <= r && r <= _n);
    assert(f(e()));
    if (r == 0) return 0;
    r += size;
    S sm = e();
    do {
      r--;
      while (r > 1 && (r % 2)) r >>= 1;
      if (!f(op(d[r], sm))) {
        while (r < size) {
          r = (2 * r + 1);
          if (f(op(d[r], sm))) {
            sm = op(d[r], sm);
            r--;
          }
        }
        return r + 1 - size;
      }
      sm = op(d[r], sm);
    } while ((r & -r) != r);
    return 0;
  }

 private:
  int _n, size, log;
  std::vector<S> d;
  void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};
}  // namespace atcoder
using namespace std;
using namespace atcoder;
using vec_int = vector<long long>;
using P = pair<long long, long long>;
using T = tuple<long long, long long, long long>;
using ll = long long;
long long charToInt(char c) {
  char zero_num = '0';
  return (long long)c - (long long)zero_num;
}
long long op(long long a, long long b) { return a + b; }
long long e() { return 0; }
signed main() {
  long long n;
  cin >> n;
  long long MOD = 998244353;
  vec_int x(n), y(n), s(n);
  for (long long i = 0; i < (long long)(n); i++)
    cin >> x.at(i) >> y.at(i) >> s.at(i);
  segtree<long long, op, e> seg(n);
  for (long long i = 0; i < n; i++) {
    long long x_pos = x.at(i);
    long long y_pos = y.at(i);
    auto iter = lower_bound(x.begin(), x.end(), y_pos);
    long long index = distance(x.begin(), iter);
    long long temp = x_pos - y_pos;
    temp %= MOD;
    temp += seg.prod(index, i);
    temp %= MOD;
    seg.set(i, temp);
  }
  long long ans = x.at(n - 1) + 1;
  for (long long i = 0; i < n; i++) {
    if (s.at(i) == 1) {
      ans += seg.get(i);
      ans %= MOD;
    }
  }
  cout << ans % MOD << endl;
  return 0;
}

### Prompt

Please formulate 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>
namespace atcoder {
namespace internal {
int ceil_pow2(int n) {
  int x = 0;
  while ((1U << x) < (unsigned int)(n)) x++;
  return x;
}
int bsf(unsigned int n) { return __builtin_ctz(n); }
}  // namespace internal
}  // namespace atcoder
namespace atcoder {
template <class S, S (*op)(S, S), S (*e)()>
struct segtree {
 public:
  segtree() : segtree(0) {}
  segtree(int n) : segtree(std::vector<S>(n, e())) {}
  segtree(const std::vector<S>& v) : _n(int(v.size())) {
    log = internal::ceil_pow2(_n);
    size = 1 << log;
    d = std::vector<S>(2 * size, e());
    for (int i = 0; i < _n; i++) d[size + i] = v[i];
    for (int i = size - 1; i >= 1; i--) {
      update(i);
    }
  }
  void set(int p, S x) {
    assert(0 <= p && p < _n);
    p += size;
    d[p] = x;
    for (int i = 1; i <= log; i++) update(p >> i);
  }
  S get(int p) {
    assert(0 <= p && p < _n);
    return d[p + size];
  }
  S prod(int l, int r) {
    assert(0 <= l && l <= r && r <= _n);
    S sml = e(), smr = e();
    l += size;
    r += size;
    while (l < r) {
      if (l & 1) sml = op(sml, d[l++]);
      if (r & 1) smr = op(d[--r], smr);
      l >>= 1;
      r >>= 1;
    }
    return op(sml, smr);
  }
  S all_prod() { return d[1]; }
  template <bool (*f)(S)>
  int max_right(int l) {
    return max_right(l, [](S x) { return f(x); });
  }
  template <class F>
  int max_right(int l, F f) {
    assert(0 <= l && l <= _n);
    assert(f(e()));
    if (l == _n) return _n;
    l += size;
    S sm = e();
    do {
      while (l % 2 == 0) l >>= 1;
      if (!f(op(sm, d[l]))) {
        while (l < size) {
          l = (2 * l);
          if (f(op(sm, d[l]))) {
            sm = op(sm, d[l]);
            l++;
          }
        }
        return l - size;
      }
      sm = op(sm, d[l]);
      l++;
    } while ((l & -l) != l);
    return _n;
  }
  template <bool (*f)(S)>
  int min_left(int r) {
    return min_left(r, [](S x) { return f(x); });
  }
  template <class F>
  int min_left(int r, F f) {
    assert(0 <= r && r <= _n);
    assert(f(e()));
    if (r == 0) return 0;
    r += size;
    S sm = e();
    do {
      r--;
      while (r > 1 && (r % 2)) r >>= 1;
      if (!f(op(d[r], sm))) {
        while (r < size) {
          r = (2 * r + 1);
          if (f(op(d[r], sm))) {
            sm = op(d[r], sm);
            r--;
          }
        }
        return r + 1 - size;
      }
      sm = op(d[r], sm);
    } while ((r & -r) != r);
    return 0;
  }

 private:
  int _n, size, log;
  std::vector<S> d;
  void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};
}  // namespace atcoder
using namespace std;
using namespace atcoder;
using vec_int = vector<long long>;
using P = pair<long long, long long>;
using T = tuple<long long, long long, long long>;
using ll = long long;
long long charToInt(char c) {
  char zero_num = '0';
  return (long long)c - (long long)zero_num;
}
long long op(long long a, long long b) { return a + b; }
long long e() { return 0; }
signed main() {
  long long n;
  cin >> n;
  long long MOD = 998244353;
  vec_int x(n), y(n), s(n);
  for (long long i = 0; i < (long long)(n); i++)
    cin >> x.at(i) >> y.at(i) >> s.at(i);
  segtree<long long, op, e> seg(n);
  for (long long i = 0; i < n; i++) {
    long long x_pos = x.at(i);
    long long y_pos = y.at(i);
    auto iter = lower_bound(x.begin(), x.end(), y_pos);
    long long index = distance(x.begin(), iter);
    long long temp = x_pos - y_pos;
    temp %= MOD;
    temp += seg.prod(index, i);
    temp %= MOD;
    seg.set(i, temp);
  }
  long long ans = x.at(n - 1) + 1;
  for (long long i = 0; i < n; i++) {
    if (s.at(i) == 1) {
      ans += seg.get(i);
      ans %= MOD;
    }
  }
  cout << ans % MOD << endl;
  return 0;
}

```