Problem: There is a chess tournament in All-Right-City. n players were invited to take part in the competition. The tournament is held by the following rules:

  1. Initially, each player plays one game with every other player. There are no ties; 
  2. After that, the organizers build a complete directed graph with players as vertices. For every pair of players there is exactly one directed edge between them: the winner of their game is the startpoint of this edge and the loser is the endpoint; 
  3. After that, the organizers build a condensation of this graph. The condensation of this graph is an acyclic complete graph, therefore it has the only Hamiltonian path which consists of strongly connected components of initial graph A1 → A2 → ... → Ak. 
  4. The players from the first component A1 are placed on the first <image> places, the players from the component A2 are placed on the next <image> places, and so on. 
  5. To determine exact place of each player in a strongly connected component, all the procedures from 1 to 5 are repeated recursively inside each component, i.e. for every i = 1, 2, ..., k players from the component Ai play games with each other again, and so on; 
  6. If a component consists of a single player, then he has no more rivals, his place is already determined and the process stops. 



The players are enumerated with integers from 1 to n. The enumeration was made using results of a previous tournament. It is known that player i wins player j (i < j) with probability p.

You need to help to organize the tournament. Find the expected value of total number of games played by all the players. 

It can be shown that the answer can be represented as <image>, where P and Q are coprime integers and <image>. Print the value of P·Q - 1 modulo 998244353.

If you are not familiar with any of the terms above, you can read about them [here](https://en.wikipedia.org/wiki/Strongly_connected_component).

Input

The first line of input contains a single integer n (2 ≤ n ≤ 2000) — the number of players.

The second line contains two integers a and b (1 ≤ a < b ≤ 100) — the numerator and the denominator of fraction <image>.

Output

In the only line print the expected value of total number of games played by all the players. Print the answer using the format above.

Examples

Input

3
1 2


Output

4


Input

3
4 6


Output

142606340


Input

4
1 2


Output

598946623

Note

In the first example the expected value is 4.

In the second example the expected value is <image>.

In the third example the expected value is <image>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
void debug(T a, T b) {
  ;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
namespace std {
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& a) {
  out << '(' << a.first << ',' << a.second << ')';
  return out;
}
}  // namespace std
long long int readL() {
  long long int res;
  scanf("%I64d", &res);
  return res;
}
void printL(long long int res) { printf("%I64d", res); }
template <long long int mod>
struct Int_ {
  unsigned x;
  unsigned mpow(Int_ a, unsigned k) {
    Int_ res = 1;
    while (k) {
      if (k & 1) res = res * a;
      a = a * a;
      k >>= 1;
    }
    return res.x;
  }
  unsigned inverse(Int_ a) { return mpow(a, mod - 2); }
  Int_() : x(0) {}
  Int_(long long sig) {
    int sigt = sig % mod;
    if (sigt < 0) sigt += mod;
    x = sigt;
  }
  unsigned get() const { return (unsigned)x; }
  Int_& operator+=(Int_ that) {
    if ((x += that.x) >= mod) x -= mod;
    return *this;
  }
  Int_& operator-=(Int_ that) {
    if ((x += mod - that.x) >= mod) x -= mod;
    return *this;
  }
  Int_& operator*=(Int_ that) {
    x = (unsigned long long)x * that.x % mod;
    return *this;
  }
  Int_& operator=(Int_ that) {
    x = that.x;
    return *this;
  }
  Int_& operator/=(Int_ that) {
    x = (unsigned long long)x * inverse(that.x) % mod;
    return *this;
  }
  bool operator==(Int_ that) const { return x == that.x; }
  bool operator!=(Int_ that) const { return x != that.x; }
  Int_ operator-() const { return Int_(0) - Int_(*this); }
  Int_ operator+(Int_ that) const { return Int_(*this) += that; }
  Int_ operator-(Int_ that) const { return Int_(*this) -= that; }
  Int_ operator*(Int_ that) const { return Int_(*this) *= that; }
  Int_ operator/(Int_ that) const { return Int_(*this) /= that; }
  Int_ inv() { return Int_(mpow(x, mod - 2)); }
};
namespace std {
template <long long int mod>
ostream& operator<<(ostream& out, const Int_<mod>& a) {
  out << a.get();
  return out;
}
template <long long int mod>
istream& operator>>(istream& in, Int_<mod>& a) {
  in >> a.x;
  return in;
}
};  // namespace std
Int_<998244353> mpow(Int_<998244353> a, long long int k) {
  Int_<998244353> res = 1;
  while (k) {
    if (k & 1) res = res * a;
    a = a * a;
    k >>= 1;
  }
  return res;
}
Int_<998244353> dp[2005][2005];
Int_<998244353> dp2[2005];
Int_<998244353> dp3[2005][2005];
Int_<998244353> dp4[2005];
int n;
int a, b;
Int_<998244353> pwX[4000005], pwY[4000005];
void showfrac(int a, int b) {
  Int_<998244353> sample = a;
  sample /= b;
  ;
  ;
}
int main() {
  cin >> n >> a >> b;
  Int_<998244353> X = a;
  X /= b;
  Int_<998244353> Y = -X + 1;
  ;
  ;
  ;
  ;
  pwX[0] = pwY[0] = 1;
  for (int i = 0; i < (n * n + 1); ++i) {
    pwX[i + 1] = pwX[i] * X;
    pwY[i + 1] = pwY[i] * Y;
  }
  dp3[1][0] = 1;
  dp3[1][1] = 1;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      dp3[i + 1][j + 1] += dp3[i][j] * pwX[(i - j)];
      dp3[i + 1][j] += dp3[i][j] * pwY[j];
    }
  }
  dp2[1] = 0;
  for (int i = 2; i <= n; ++i) {
    for (int j = (1); j < (i); ++j) dp[i][j] = dp3[i][j];
    for (int j = (1); j < (i); ++j) {
      Int_<998244353> T = dp4[j];
      dp[i][j] -= dp[i][j] * T;
    }
    Int_<998244353> tot = 0;
    for (int j = (1); j < (i); ++j) {
      tot += dp[i][j];
    }
    dp4[i] = tot;
    for (int j = (1); j < (i); ++j) {
      dp2[i] += dp[i][j] * ((dp2[i - j] - (i - j) * (i - j - 1) / 2) + dp2[j]);
    }
    dp2[i] += i * (i - 1) / 2;
    dp2[i] /= tot;
  }
  cout << dp2[n] << endl;
  return 0;
}