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;
typedef long long lint;

const int MAX_N = (int)2e5 + 5;
const lint MOD = (lint)1e9 + 7;
lint fact[MAX_N], fact_inv[MAX_N], ap[MAX_N], bp[MAX_N], inv_ab[MAX_N], inv[MAX_N];

void mod_add(lint &a, lint b){
    a = (a+b) % MOD;
}

lint mod_pow(lint n, lint x){
    lint res = 1;
    while(x){
        if(x&1) res = res * n % MOD;
        x >>= 1, n = n * n % MOD;
    }
    return res;
}

lint comb(int n, int r){
    return fact[n] * fact_inv[r] % MOD * fact_inv[n-r] % MOD;
}

int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int N; cin >> N;
    lint A, B, C; cin >> A >> B >> C;
    fact[0] = ap[0] = bp[0] = inv_ab[0] = 1;
    for(int i=1;i<=(int)2e5;i++){
        fact[i] = fact[i-1] * i % MOD;
        ap[i] = ap[i-1] * A % MOD;
        bp[i] = bp[i-1] * B % MOD;
        inv_ab[i] = mod_pow(A+B, MOD-i-1);
    }
    fact_inv[2*N] = mod_pow(fact[2*N], MOD-2);
    for(int i=2*N-1;0<=i;i--) fact_inv[i] = fact_inv[i+1] * (i+1) % MOD;
    for(int i=1;i<=100;i++) inv[i] = mod_pow(i, MOD-2);

    lint ans = 0;
    for(int i=N;i<2*N;i++){
        mod_add(ans, comb(i-1, N-1)*((ap[N]*bp[i-N]%MOD+ap[i-N]*bp[N]%MOD)%MOD)%MOD*i%MOD*inv_ab[i]%MOD*inv[100-C]%MOD*100%MOD);
    }
    cout << ans << endl;
    return 0;
}