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>

const int moder = (int) 1e9 + 7;
const int N = 1000010;

int fac[N], inv[N], invf[N];

int powermod(int a, int exp){
    int ret = 1;
    for ( ; exp > 0; exp >>= 1){
        if (exp & 1){
            ret = 1ll * ret * a % moder;
        }
        a = 1ll * a * a % moder;
    }
    return ret;
}

int comb(int a, int b){
    return 1ll * fac[a] * invf[b] % moder * invf[a - b] % moder;
}

int main(){
    fac[0] = invf[0] = 1;
    for (int i = 1; i < N; ++ i){
        fac[i] = 1ll * fac[i - 1] * i % moder;
        inv[i] = i == 1 ? 1 : moder - 1ll * (moder / i) * inv[moder % i] % moder;
        invf[i] = 1ll * invf[i - 1] * inv[i] % moder;
    }
    int n, a, b, c;
    scanf("%d%d%d%d", &n, &a, &b, &c);
    int sum = 0;
    int aa = 1ll * a * inv[a + b] % moder;
    int bb = 1ll * b * inv[a + b] % moder;
    for (int i = n; i <= 2 * n - 1; ++ i){
        int coe = (1ll * powermod(aa, n) * powermod(bb, i - n) + 1ll * powermod(aa, i - n) * powermod(bb, n)) % moder;
        sum = (sum + 1ll * coe * comb(i - 1, n - 1) % moder * i) % moder;
    }
    sum = 1ll * sum * 100 % moder * inv[100 - c] % moder;
    printf("%d\n", sum);
    return 0;
}