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 <bits/stdc++.h>

using namespace std;
using ll = long long;

ll mod = 1000000007;

const int MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];

void Cinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = mod - inv[mod%i] * (mod / i) % mod;
    finv[i] = finv[i - 1] * inv[i] % mod;
  }
}

ll nCk(ll n, ll k) {
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}

ll mpow(ll a, ll b)
{
  if (b == 0)return 1;
  ll z = mpow(a, b / 2);
  z = z*z%mod;
  if (b % 2 == 1)z = z*a%mod;
  return z;
}

int main() {
  Cinit();
  ll A, B, C, N;
  cin >> N >> A >> B >> C;
  ll ka = A * inv[A + B] % mod;
  ll kb = B * inv[A + B] % mod;
  ll r = 0;
  for (int i = 0; i <= N - 1; i++)
  {
    ll sa = nCk(N - 1 + i, N - 1)*ka%mod*mpow(ka, N - 1) % mod*mpow(kb, i);
    ll sb = nCk(N - 1 + i, N - 1)*kb%mod*mpow(kb, N - 1) % mod*mpow(ka, i);
    r = (r + (sa + sb) % mod*(N + i)) % mod;
  }
  cout << r * 100 % mod*inv[A + B] % mod << endl;
}
