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>
#include<iostream>
#include<fstream>
#define ll long long
using namespace std;

vector<ll> fct, invfct;

ll mod_pow(ll n, ll p, ll mod){
    // return (n^p) % mod
    if(p==0) return 1;
    ll res = mod_pow(n*n%mod, p/2, mod);
    if(p%2==1) res = res * n % mod;
    return res;
}

int main(){
    ll N, A, B, C;
    cin >> N >> A >> B >> C;
    ll mod = 1e9+7;

    fct = vector<ll>(2*N+1,1);
    for(ll i=1; i<=2*N; i++){
        fct[i] = i * fct[i-1] % mod;
    }
    invfct = vector<ll>(2*N+1,1);
    invfct[2*N] = mod_pow(fct[2*N], mod-2, mod) % mod;
    for(ll i=2*N-1; i>=1; i--){
        invfct[i] = (i+1)*invfct[i+1] % mod;
    }
    vector<ll> invAB(2*N+1,1);
    invAB[1] = mod_pow(A+B, mod-2, mod);
    for(int i=1; i<=2*N; i++){
        invAB[i] = invAB[i-1]*invAB[1] % mod;
    }
    vector<ll> powA(2*N+1,1), powB(2*N+1,1);
    powA[1] = A, powB[1] = B;
    for(int i=1; i<=2*N; i++){
        powA[i] = powA[i-1] * A % mod;
        powB[i] = powB[i-1] * B % mod;
    }

    ll ans = 0;
    for(int M=N; M<=2*N-1; M++){
        ll tmp = fct[M-1]*invfct[N-1] % mod;
        tmp = tmp * invfct[M-N] % mod;
        tmp = tmp * M % mod;
        tmp = tmp * (powA[N]*powB[M-N]%mod + powA[M-N]*powB[N]%mod) % mod;
        tmp = tmp * invAB[M] % mod;
        ans = (ans + tmp)%mod;
    }
    ans = ans * 100 % mod;
    ans = ans * mod_pow(100-C, mod-2, mod) % mod;
    cout << ans << endl;
}
