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 <iostream>
#include <vector>
using namespace std;
long long const mod = 1000000007;

long long mod_pow(long long a, long long b){
    long long res = 1;
    a %= mod;
    while(b > 0){
        if(b%2 == 1) res = (res * a) % mod;
        a = (a * a) % mod;
        b /= 2;
    }
    return res;
}

long long mod_div(long long a){
    return mod_pow(a, mod-2);
}

int main(void){
    long long n, a, b, c, ab, an, ak, bn, bk, kcn, ans = 0;
    cin >> n >> a >> b >> c;
    ab = mod_div(a+b);
    a = a*ab % mod;
    b = b*ab % mod;
    an = mod_pow(a, n);
    bn = mod_pow(b, n);
    ak = bk = kcn = 1;
    for(long long k = n; k < n*2; k++){
        ans += (an * bk + bn * ak) % mod * kcn % mod * k % mod;
        ak = ak*a % mod;
        bk = bk*b % mod;
        kcn = kcn*k % mod * mod_div(k-n+1) % mod;
    }
    ans = ans * 100 % mod * mod_div(100-c) % mod;
    cout << ans << endl;
}
