#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
using namespace std;
template <class T>
struct LazySegmentTree : public T {
  using S = typename T::S;
  using F = typename T::F;
  LazySegmentTree(int n) : d(4 * n, T::e()), lz(4 * n, T::id()) {
    for (; sz < n; sz *= 2)
      ;
  }
  void apply(int l, int r, F f) { apply(l, r, f, 0, 0, sz); }
  S prod(int l, int r) { return prod(l, r, 0, 0, sz); }

 private:
  int sz = 1;
  vector<S> d;
  vector<F> lz;
  void propagate(int x, int lx, int rx) {
    if (rx - lx == 1) return;
    lz[2 * x + 1] = T::composition(lz[x], lz[2 * x + 1]);
    lz[2 * x + 2] = T::composition(lz[x], lz[2 * x + 2]);
    d[2 * x + 1] = T::mapping(lz[x], d[2 * x + 1], (rx - lx) / 2);
    d[2 * x + 2] = T::mapping(lz[x], d[2 * x + 2], (rx - lx) / 2);
    lz[x] = T::id();
  }
  void apply(int l, int r, F f, int x, int lx, int rx) {
    propagate(x, lx, rx);
    if (rx <= l or r <= lx) return;
    if (l <= lx and rx <= r) {
      lz[x] = T::composition(f, lz[x]);
      d[x] = T::mapping(f, d[x], rx - lx);
      return;
    }
    apply(l, r, f, 2 * x + 1, lx, (lx + rx) / 2);
    apply(l, r, f, 2 * x + 2, (lx + rx) / 2, rx);
    d[x] = T::op(d[2 * x + 1], d[2 * x + 2]);
  }
  S prod(int l, int r, int x, int lx, int rx) {
    propagate(x, lx, rx);
    if (rx <= l or r <= lx) return T::e();
    if (l <= lx and rx <= r) return d[x];
    return T::op(prod(l, r, 2 * x + 1, lx, (lx + rx) / 2),
                 prod(l, r, 2 * x + 2, (lx + rx) / 2, rx));
  }
};
const int MOD = 998244353;
struct Node {
  using S = long long;
  using F = long long;
  S op(S a, S b) { return (a + b) % MOD; }
  S e() { return 0; }
  S mapping(F f, S x, int len) { return (x + f * len) % MOD; }
  F composition(F f, F g) { return (f + g) % MOD; }
  F id() { return 0; }
};
struct Portal {
  int x, y;
  bool s;
};
void solve() {
  int n;
  cin >> n;
  vector<Portal> portals(n);
  for (auto &portal : portals) cin >> portal.x >> portal.y >> portal.s;
  LazySegmentTree<Node> st(n);
  for (int i = n - 1; i >= 0; --i) {
    if (portals[i].s) {
      st.apply(i, i + 1, 1);
    }
    int lo = -1, hi = i;
    while (lo + 1 < hi) {
      int mid = (lo + hi) / 2;
      if (portals[mid].x >= portals[i].y)
        hi = mid;
      else
        lo = mid;
    }
    st.apply(hi, i, st.prod(i, i + 1));
  }
  long long ans = portals[n - 1].x + 1;
  for (int i = 0; i < n; ++i)
    ans = (ans + st.prod(i, i + 1) * (portals[i].x - portals[i].y)) % MOD;
  cout << ans << "\n";
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  solve();
}

### Prompt

Please create a solution in cpp 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
using namespace std;
template <class T>
struct LazySegmentTree : public T {
  using S = typename T::S;
  using F = typename T::F;
  LazySegmentTree(int n) : d(4 * n, T::e()), lz(4 * n, T::id()) {
    for (; sz < n; sz *= 2)
      ;
  }
  void apply(int l, int r, F f) { apply(l, r, f, 0, 0, sz); }
  S prod(int l, int r) { return prod(l, r, 0, 0, sz); }

 private:
  int sz = 1;
  vector<S> d;
  vector<F> lz;
  void propagate(int x, int lx, int rx) {
    if (rx - lx == 1) return;
    lz[2 * x + 1] = T::composition(lz[x], lz[2 * x + 1]);
    lz[2 * x + 2] = T::composition(lz[x], lz[2 * x + 2]);
    d[2 * x + 1] = T::mapping(lz[x], d[2 * x + 1], (rx - lx) / 2);
    d[2 * x + 2] = T::mapping(lz[x], d[2 * x + 2], (rx - lx) / 2);
    lz[x] = T::id();
  }
  void apply(int l, int r, F f, int x, int lx, int rx) {
    propagate(x, lx, rx);
    if (rx <= l or r <= lx) return;
    if (l <= lx and rx <= r) {
      lz[x] = T::composition(f, lz[x]);
      d[x] = T::mapping(f, d[x], rx - lx);
      return;
    }
    apply(l, r, f, 2 * x + 1, lx, (lx + rx) / 2);
    apply(l, r, f, 2 * x + 2, (lx + rx) / 2, rx);
    d[x] = T::op(d[2 * x + 1], d[2 * x + 2]);
  }
  S prod(int l, int r, int x, int lx, int rx) {
    propagate(x, lx, rx);
    if (rx <= l or r <= lx) return T::e();
    if (l <= lx and rx <= r) return d[x];
    return T::op(prod(l, r, 2 * x + 1, lx, (lx + rx) / 2),
                 prod(l, r, 2 * x + 2, (lx + rx) / 2, rx));
  }
};
const int MOD = 998244353;
struct Node {
  using S = long long;
  using F = long long;
  S op(S a, S b) { return (a + b) % MOD; }
  S e() { return 0; }
  S mapping(F f, S x, int len) { return (x + f * len) % MOD; }
  F composition(F f, F g) { return (f + g) % MOD; }
  F id() { return 0; }
};
struct Portal {
  int x, y;
  bool s;
};
void solve() {
  int n;
  cin >> n;
  vector<Portal> portals(n);
  for (auto &portal : portals) cin >> portal.x >> portal.y >> portal.s;
  LazySegmentTree<Node> st(n);
  for (int i = n - 1; i >= 0; --i) {
    if (portals[i].s) {
      st.apply(i, i + 1, 1);
    }
    int lo = -1, hi = i;
    while (lo + 1 < hi) {
      int mid = (lo + hi) / 2;
      if (portals[mid].x >= portals[i].y)
        hi = mid;
      else
        lo = mid;
    }
    st.apply(hi, i, st.prod(i, i + 1));
  }
  long long ans = portals[n - 1].x + 1;
  for (int i = 0; i < n; ++i)
    ans = (ans + st.prod(i, i + 1) * (portals[i].x - portals[i].y)) % MOD;
  cout << ans << "\n";
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  solve();
}

```