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 <algorithm>
#include <vector>
#include <math.h>
using namespace std;
typedef long long ll;

ll inf = 1e9+7;

ll pw(ll x, ll y) {
    if (y == 0) return 1;
    else if (y == 1) return x % inf;
    else if (y % 2 == 0) return pw(x, y / 2) * pw(x, y / 2) % inf;
    else return pw(x, y / 2) * pw(x, y / 2) % inf * x % inf;
}

ll dv(ll x, ll y) {
    return x * pw(y, inf - 2) % inf;
}

int main(void){
    ll n, a, b, c;
    cin >> n >> a >> b >> c;
    
    ll ans = 0, comb = 1;
    for (int i = 0; i < n; i++) {
        ans += (comb * pw(dv(a, 100 - c), n) % inf) * pw(dv(b, 100 - c), i) % inf * (n + i) % inf;
        ans += (comb * pw(dv(b, 100 - c), n) % inf) * pw(dv(a, 100 - c), i) % inf * (n + i) % inf;
        comb *= (n + i) * dv(1, i + 1) % inf;
        ans %= inf;
        comb %= inf;
    }
    
    cout << ans % inf * dv(100, 100 - c) % inf << endl;
    
}
