Takahashi and Aoki will play a game. They will repeatedly play it until one of them have N wins in total.

When they play the game once, Takahashi wins with probability A %, Aoki wins with probability B %, and the game ends in a draw (that is, nobody wins) with probability C %. Find the expected number of games that will be played, and print it as follows.

We can represent the expected value as P/Q with coprime integers P and Q. Print the integer R between 0 and 10^9+6 (inclusive) such that R \times Q \equiv P\pmod {10^9+7}. (Such an integer R always uniquely exists under the constraints of this problem.)

Constraints

* 1 \leq N \leq 100000
* 0 \leq A,B,C \leq 100
* 1 \leq A+B
* A+B+C=100
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N A B C


Output

Print the expected number of games that will be played, in the manner specified in the statement.

Examples

Input

1 25 25 50


Output

2


Input

4 50 50 0


Output

312500008


Input

1 100 0 0


Output

1


Input

100000 31 41 28


Output

104136146
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <functional>

using namespace std;
typedef long long int ll;

#define repi(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,a) repi(i,0,a)
#define rrep(i,a) for(ll i=a-1;i>=0;i--)
#define MOD 1000000007

ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

ll modpow(ll a, ll n, ll mod){
  ll tmp = a, val = 1;
  while(n){
    if(n % 2) val = val * tmp % mod;
    tmp = tmp * tmp % mod;
    n >>= 1;
  }
  return val;
}

int main(){
  // P(i, j) = (A^n * B^i) * (n+i)!/i!n!
  // P(i+1) = P(i) * (B / (A + B) * (n+i+1) / (i+1) + 1) * (n+i+1) / (n+i)
  ll N, A, B, C;
  cin >> N >> A >> B >> C;
  ll Pa = modpow(A, N, MOD), Pb = modpow(B, N, MOD);
  ll P = (Pa + Pb) * N % MOD;
  ll Q = modpow(A+B, N, MOD);
  rep(i, N-1){
    Pa = Pa * B * (N + i) % MOD;
    Pb = Pb * A * (N + i) % MOD;
    Q = Q * (A + B) * (i + 1) % MOD;
    P = (P * (A + B) * (i + 1) + (Pa + Pb) * (N + i + 1)) % MOD;
    // printf("Pa: %lld Pb: %lld P: %lld Q: %lld\n", Pa, Pb, P, Q);
  }
  P = P * 100 % MOD;
  Q = Q * (100 - C) % MOD;
  ll R = P * modinv(Q, MOD) % MOD;

  cout << R << endl;
  return 0;
}
