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 ll long long


ll mod = 1e9+7;

ll modpower(ll a, ll r){ //a^r
    ll x = 1;
    while (r > 0) {
        if (r & 1) x = x * a % mod;
        r >>= 1;
        a = a * a % mod;
    }
    return x;
}

ll modinv(ll a){
    return modpower(a, mod-2);
}


int main(){
    ll N, A, B, C;
    cin >> N >> A >> B >> C;
    A = (A*modinv(100))%mod;
    B = (B*modinv(100))%mod;
    C = (C*modinv(100))%mod;
    ll ans = 0;
    ll I1 = modpower(A, N) * modinv(modpower(mod+1-C, N+1)) % mod;
    ll I2 = modpower(B, N) * modinv(modpower(mod+1-C, N+1)) % mod;
    ll J1 = A*modinv(mod+1-C) % mod;
    ll J2 = B*modinv(mod+1-C) % mod;
    ll T = N;
    
    ans += I1*T%mod;
    for (int K = 1; K <= N-1; K++) {
        T = (T*(N+K)%mod)*modinv(K)%mod*J2%mod;
        ans += I1*T%mod;
    }
    
    
    T = N;
    ans += I2*T%mod;
    for (int K = 1; K <= N-1; K++) {
        T = (T*(N+K)%mod)*modinv(K)%mod*J1%mod;
        ans += I2*T%mod;
    }
    cout << ans%mod << endl;
    return 0;
}
