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;

long long F[200001], AA[200001], BB[200001], D[200001], MOD=1000000007;

long long Power(long long a, long long k)
{
    long long t=1, S=a, Su=1;
    while (k > 0)
    {
        if (k % (2*t) != 0)
        {
            k -= t;
            Su *= S;
            Su %= MOD;
        }
        t *= 2;
        S *= S;
        S %= MOD;
    }
    return Su;
}

int main()
{
    long long N, A, B, C, P=0, Q=1, S1, S2, S;
    cin >> N >> A >> B >> C;
    AA[0] = 1;
    BB[0] = 1;
    F[0] = 1;
    D[0] = 1;
    for (int i = 1; i <= 2*N; i++)
    {
        AA[i] = AA[i-1] * A;
        BB[i] = BB[i-1] * B;
        AA[i] %= MOD;
        BB[i] %= MOD;
        F[i] = F[i-1] * i;
        F[i] %= MOD;
        D[i] = D[i-1] * (100-C);
        D[i] %= MOD;
    }
    for (int i = N; i <= 2*N-1; i++)
    {
        S1 = F[i-1];
        S2 = F[N-1];
        S1 *= i;
        S1 %= MOD;
        S = AA[N]*BB[i-N] + AA[i-N]*BB[N];
        S %= MOD;
        S1 *= S;
        S1 %= MOD;
        S1 *= 100;
        S1 %= MOD;
        S2 *= F[i-N];
        S2 %= MOD;
        S2 *= D[i+1];
        S2 %= MOD;
        P = (P*S2 + S1*Q) % MOD;
        Q = (S2*Q) % MOD;
    }
    S = (P * Power(Q, MOD-2)) % MOD;
    cout << S;
    return 0;
}
