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;
const int max_n = 100111, inf = 2000111222;
const int mod = 998244353;
int mul(int x, int y) { return (1LL * x * y) % mod; }
int power(int a, int b) {
  if (b == 0) {
    return 1 % mod;
  }
  if (b % 2 == 0) {
    return power(mul(a, a), b / 2);
  }
  return mul(a, power(a, b - 1));
}
int inv(int x) { return power(x, mod - 2); }
struct frac {
  int a, b;
  frac() : a(0), b(1) {}
  frac(int a, int b) : a(a), b(b) {}
  int get_val() const { return mul(a, inv(b)); }
  bool operator<(const frac &f) const { return 1LL * a * f.b < 1LL * f.a * b; }
  bool operator==(const frac &f) const {
    return 1LL * a * f.b == 1LL * f.a * b;
  }
  bool operator<=(const frac &f) const {
    return 1LL * a * f.b <= 1LL * f.a * b;
  }
  friend ostream &operator<<(ostream &os, const frac &f) {
    os << f.a << "/" << f.b;
    return os;
  }
};
vector<pair<int, int>> upd;
struct state {
  vector<int> v[2][2];
  vector<int> diff;
};
const int inv100 = inv(100);
int n, x[max_n], v[max_n], pr[max_n], pl[max_n];
int nums[max_n][2][2];
int buf[4 * max_n];
vector<frac> all;
frac get_frac(int pos, int tp2, int tp3) {
  frac upd(inf, 1);
  if (tp2 == 1 && tp3 == 0) {
    upd = {x[pos + 1] - x[pos], v[pos] + v[pos + 1]};
  } else if (tp2 == 0 && tp3 == 0 && v[pos] < v[pos + 1]) {
    upd = {x[pos + 1] - x[pos], v[pos + 1] - v[pos]};
  } else if (tp2 == 1 && tp3 == 1 && v[pos] > v[pos + 1]) {
    upd = {x[pos + 1] - x[pos], v[pos] - v[pos + 1]};
  }
  return upd;
}
state merge(const state &a, const state &b, int pos) {
  state res;
  res.diff.resize(a.diff.size() + b.diff.size());
  merge(a.diff.begin(), a.diff.end(), b.diff.begin(), b.diff.end(),
        res.diff.begin());
  for (int tp2 = 0; tp2 < 2; ++tp2) {
    for (int tp3 = 0; tp3 < 2; ++tp3) {
      const int nval = nums[pos][tp2][tp3];
      res.diff.insert(lower_bound(res.diff.begin(), res.diff.end(), nval),
                      nval);
    }
  }
  assert(is_sorted(res.diff.begin(), res.diff.end()));
  res.diff.erase(unique(res.diff.begin(), res.diff.end()), res.diff.end());
  for (int tp1 = 0; tp1 < 2; ++tp1) {
    for (int tp4 = 0; tp4 < 2; ++tp4) {
      for (int tp2 = 0; tp2 < 2; ++tp2) {
        for (int tp3 = 0; tp3 < 2; ++tp3) {
          if (a.v[tp1][tp2].empty() || b.v[tp3][tp4].empty()) {
            continue;
          }
          const vector<int> &v1 = a.v[tp1][tp2];
          const vector<int> &v2 = b.v[tp3][tp4];
          const int nval = nums[pos][tp2][tp3];
          int cur = 0;
          int sump = 0;
          for (int p : v2) {
            sump += p;
            sump %= mod;
          }
          for (int i = 0; i < a.diff.size(); ++i) {
            const int val = min(nval, a.diff[i]);
            while (cur < b.diff.size() && b.diff[cur] < val) {
              sump += mod - v2[cur];
              sump %= mod;
              ++cur;
            }
            buf[val] += mul(v1[i], sump);
            buf[val] %= mod;
          }
          sump = cur = 0;
          for (int p : v1) {
            sump += p;
            sump %= mod;
          }
          for (int i = 0; i < b.diff.size(); ++i) {
            const int val = b.diff[i];
            while (cur < a.diff.size() && min(a.diff[cur], nval) <= val) {
              sump += mod - v1[cur];
              sump %= mod;
              ++cur;
            }
            buf[val] += mul(v2[i], sump);
            buf[val] %= mod;
          }
        }
      }
      for (int x : res.diff) {
        res.v[tp1][tp4].push_back(buf[x]);
        buf[x] = 0;
      }
    }
  }
  return res;
}
state solve(int l, int r) {
  if (l == r) {
    state res;
    res.v[0][0].push_back(pl[l]);
    res.v[1][1].push_back(pr[l]);
    res.diff.push_back(all.size() - 1);
    return res;
  }
  int mid = (l + r) / 2;
  state a = solve(l, mid);
  state b = solve(mid + 1, r);
  return merge(a, b, mid);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d%d", &x[i], &v[i], &pr[i]);
  }
  for (int i = 0; i < n; ++i) {
    pr[i] = mul(pr[i], inv100);
    pl[i] = (1 + mod - pr[i]) % mod;
  }
  for (int i = 0; i + 1 < n; ++i) {
    for (int tp2 = 0; tp2 < 2; ++tp2) {
      for (int tp3 = 0; tp3 < 2; ++tp3) {
        all.push_back(get_frac(i, tp2, tp3));
      }
    }
  }
  all.push_back({inf, 1});
  sort(all.begin(), all.end());
  all.erase(unique(all.begin(), all.end()), all.end());
  for (int i = 0; i + 1 < n; ++i) {
    for (int tp2 = 0; tp2 < 2; ++tp2) {
      for (int tp3 = 0; tp3 < 2; ++tp3) {
        nums[i][tp2][tp3] =
            lower_bound(all.begin(), all.end(), get_frac(i, tp2, tp3)) -
            all.begin();
      }
    }
  }
  state s = solve(0, n - 1);
  int ans = 0;
  for (int f1 = 0; f1 < 2; ++f1) {
    for (int f2 = 0; f2 < 2; ++f2) {
      for (int i = 0; i < s.diff.size() && i < s.v[f1][f2].size(); ++i) {
        if (all[s.diff[i]] == frac(inf, 1)) {
          continue;
        }
        ans += mul(all[s.diff[i]].get_val(), s.v[f1][f2][i]);
        ans %= mod;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}