Problem: An infinitely long Line Chillland Collider (LCC) was built in Chillland. There are n pipes with coordinates x_i that are connected to LCC. When the experiment starts at time 0, i-th proton flies from the i-th pipe with speed v_i. It flies to the right with probability p_i and flies to the left with probability (1 - p_i). The duration of the experiment is determined as the time of the first collision of any two protons. In case there is no collision, the duration of the experiment is considered to be zero.

Find the expected value of the duration of the experiment.

<image>Illustration for the first example

Input

The first line of input contains one integer n — the number of pipes (1 ≤ n ≤ 10^5). Each of the following n lines contains three integers x_i, v_i, p_i — the coordinate of the i-th pipe, the speed of the i-th proton and the probability that the i-th proton flies to the right in percentage points (-10^9 ≤ x_i ≤ 10^9, 1 ≤ v ≤ 10^6, 0 ≤ p_i ≤ 100). It is guaranteed that all x_i are distinct and sorted in increasing order.

Output

It's possible to prove that the answer can always be represented as a fraction P/Q, where P is an integer and Q is a natural number not divisible by 998 244 353. In this case, print P ⋅ Q^{-1} modulo 998 244 353.

Examples

Input


2
1 1 100
3 1 0


Output


1


Input


3
7 10 0
9 4 86
14 5 100


Output


0


Input


4
6 4 50
11 25 50
13 16 50
15 8 50


Output


150902884

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
int read() {
  char c;
  while ((c = getchar()) < '-')
    ;
  if (c == '-') {
    int x = (c = getchar()) - '0';
    while ((c = getchar()) >= '0') x = x * 10 + c - '0';
    return -x;
  }
  int x = c - '0';
  while ((c = getchar()) >= '0') x = x * 10 + c - '0';
  return x;
}
const int N = 2e5 + 5, D = 998244353;
int64_t mi(int64_t x, int y = D - 2) {
  int64_t ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % D;
    x = x * x % D;
    y >>= 1;
  }
  return ans;
}
void operator+=(pair<int64_t, int64_t> &a, const pair<int64_t, int64_t> &b) {
  (a.first += b.first) %= D;
  (a.second += b.second) %= D;
}
void mul(int64_t &a0, int64_t &a1, int64_t m[2][2]) {
  int64_t b0 = (m[0][0] * a0 + m[0][1] * a1) % D,
          b1 = (m[1][0] * a0 + m[1][1] * a1) % D;
  a0 = b0;
  a1 = b1;
}
int64_t m[2][2], m1[2][2];
namespace SEG {
const int64_t T = N * 4;
int64_t tag[T][2][2], sum[T][2];
int n;
void mul(int k, int64_t m[2][2]) {
  for (int i = 0; i <= 1; ++i) {
    ::mul(tag[k][0][i], tag[k][1][i], m);
  }
  ::mul(sum[k][0], sum[k][1], m);
}
void push_down(int k) {
  if ((tag[k][0][0] - 1) % D || tag[k][0][1] || tag[k][1][0] ||
      (tag[k][1][1] - 1) % D) {
    mul((k * 2), tag[k]);
    mul(((k * 2) + 1), tag[k]);
    for (int i = 0; i <= 1; ++i)
      for (int j = 0; j <= 1; ++j) tag[k][i][j] = i == j;
  }
}
void push_up(int k) {
  for (int i = 0; i <= 1; ++i)
    sum[k][i] = (sum[(k * 2)][i] + sum[((k * 2) + 1)][i]) % D;
}
void add(int x, const pair<int64_t, int64_t> &a) {
  int k = 1, l = 1, r = n;
  while (1) {
    (sum[k][0] += a.first) %= D;
    (sum[k][1] += a.second) %= D;
    if (l == r) break;
    push_down(k);
    if (x <= ((l + r) / 2)) {
      k = (k * 2);
      r = ((l + r) / 2);
    } else {
      k = ((k * 2) + 1);
      l = ((l + r) / 2) + 1;
    }
  }
}
pair<int64_t, int64_t> query(int x) {
  int k = 1, l = 1, r = n;
  pair<int64_t, int64_t> ans = pair<int64_t, int64_t>(0, 0);
  while (l != r) {
    push_down(k);
    if (x <= ((l + r) / 2)) {
      ans +=
          pair<int64_t, int64_t>(sum[((k * 2) + 1)][0], sum[((k * 2) + 1)][1]);
      k = (k * 2);
      r = ((l + r) / 2);
    } else {
      k = ((k * 2) + 1);
      l = ((l + r) / 2) + 1;
    }
  }
  ans += pair<int64_t, int64_t>(sum[k][0], sum[k][1]);
  return ans;
}
int ql, qr;
void modify(int k, int l, int r) {
  if (ql > r || qr < l) return;
  if (ql <= l && qr >= r) {
    mul(k, m);
    return;
  }
  push_down(k);
  modify((k * 2), l, ((l + r) / 2));
  modify(((k * 2) + 1), ((l + r) / 2) + 1, r);
  push_up(k);
}
void modify(int l, int r) {
  ql = l;
  qr = r;
  modify(1, 1, n);
}
};  // namespace SEG
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return 1LL * a.first * b.second < 1LL * a.second * b.first;
}
pair<int, int> q[N];
int t;
int main() {
  int n = read();
  static int x[N], v[N], p[N];
  for (int i = 1; i <= n; ++i) {
    x[i] = read();
    v[i] = read();
    p[i] = read() * mi(100) % D;
  }
  t = 1;
  q[1] = pair<int, int>(1, 0);
  for (int i = 2; i <= n; ++i) {
    q[++t] = pair<int, int>(x[i] - x[i - 1], v[i] + v[i - 1]);
    if (v[i] != v[i - 1])
      q[++t] = pair<int, int>(x[i] - x[i - 1], abs(v[i] - v[i - 1]));
  }
  sort(q + 1, q + t + 1, cmp);
  SEG::n = t;
  SEG::add(t, pair<int64_t, int64_t>(1 - p[1], p[1]));
  for (int i = 2; i <= n; ++i) {
    int j0 =
        lower_bound(q + 1, q + t,
                    pair<int, int>(x[i] - x[i - 1], v[i] + v[i - 1]), cmp) -
        q;
    int j1 = lower_bound(q + 1, q + t,
                         pair<int, int>(x[i] - x[i - 1], abs(v[i] - v[i - 1])),
                         cmp) -
             q;
    pair<int64_t, int64_t> s0 = SEG::query(j0), s1 = SEG::query(j1);
    int64_t np = p[i];
    for (int i = 0; i <= 1; ++i)
      for (int j = 0; j <= 1; ++j) m[i][j] = (i == 0) ? 1 - np : np;
    SEG::modify(1, j0 - 1);
    for (int i = 0; i <= 1; ++i)
      for (int j = 0; j <= 1; ++j)
        if (j == 1 && i == 0) {
          m1[i][j] = m[i][j];
          m[i][j] = 0;
        } else
          m1[i][j] = 0;
    mul(s0.first, s0.second, m1);
    SEG::modify(j0, j1 - 1);
    SEG::add(j0, s0);
    int ni = v[i] > v[i - 1] ? 0 : 1;
    for (int i = 0; i <= 1; ++i)
      for (int j = 0; j <= 1; ++j)
        if (i == j && (i == ni)) {
          m1[i][j] = m[i][j];
          m[i][j] = 0;
        } else
          m1[i][j] = 0;
    mul(s1.first, s1.second, m1);
    SEG::modify(j1, t);
    SEG::add(j1, s1);
  }
  auto S = [&](const pair<int64_t, int64_t> &a) -> int64_t {
    return (a.first + a.second) % D;
  };
  int64_t ans = 0;
  for (int i = 1; i <= t - 1; ++i)
    (ans += (S(SEG::query(i)) - S(SEG::query(i + 1))) * q[i].first % D *
            mi(q[i].second)) %= D;
  cout << (ans + D) % D << endl;
}