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 <iostream>
#include <cmath>

using namespace std;

typedef long long ll;

ll fac[300005];
ll rfac[300005];
const ll mod = 1e9 + 7;
ll ans;
ll p(ll a, ll x) {
  ll res = 1;
  while(x > 0) {
    if(x & 1) res = res * a % mod;
    a = a * a % mod;
    x >>= 1;
  }
  return res;
}

ll nCr(ll n, ll r) {
  return fac[n] * (rfac[r] * rfac[n-r] % mod) % mod;
}

int main() {
  ll n,a,b,c;
  cin >> n >> a >> b >> c;
  fac[0] = 1;
  rfac[0] = 1;
  for(ll i = 1; i <= 2 * n; i++) {
    fac[i] = fac[i-1] * i % mod;
    rfac[i] = p(fac[i],mod-2);
  }

  
  ll C = 1;
  if(c != 0) {
    C = (ll)100 * p(100-c,mod-2) % mod;
  }

  for(ll i = n; i <= 2 * n - 1; i++) {
    ll com = nCr(i - 1,n - 1);
    ans = (ans + com * i % mod * C % mod * p(a,n) % mod * p(b,i-n) % mod * p(p(a+b,i),mod-2) % mod) % mod;
    ans = (ans + com * i % mod * C % mod * p(b,n) % mod * p(a,i-n) % mod * p(p(a+b,i),mod-2) % mod) % mod;
  }

  cout << ans % mod<< endl;
}

  
