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;
#define int long long

const int MAX = 1010101;
const int MOD = 1e9 + 7;

int N, A, B, C;
int fac[MAX], finv[MAX], inv[MAX];

void CombInit() {
  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;
  }
}

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

int mod_pow(int x, int n, int m = MOD) {
  if (n == 0)
    return 1;
  int res = mod_pow(x * x % m, n / 2, m);
  if (n % 2 == 1)
    res = res * x % m;
  return res;
}

signed main() {
  cin >> N >> A >> B >> C;

  CombInit();

  int ans = 0;
  for (int M = N; M < 2 * N; M++) {
    int X = inv[A + B];
    int tmp = (mod_pow(A * X % MOD, N) * mod_pow(B * X % MOD, M - N)) % MOD;
    tmp += (mod_pow(A * X % MOD, M - N) * mod_pow(B * X % MOD, N)) % MOD;
    tmp %= MOD;

    tmp *= nCk(M - 1, N - 1);
    tmp %= MOD;

    int tmp2 = (M * 100 % MOD) * inv[100 - C] % MOD;

    ans += tmp * tmp2;
    ans %= MOD;
  }

  cout << ans << endl;
  return 0;
}
