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>

using namespace std;

const int Z = 1e9+7;

int fp(int a, int x) {
    if (x == 0) {
        return 1;
    }
    if (x % 2 == 1) {
        return 1ll * a * fp(a, x - 1) % Z;
    }
    int t = fp(a, x / 2);
    return 1ll * t * t % Z;
}

int inv(int a) {
    return fp(a, Z - 2);
}

int main() {
    int n, a, b, c;
    cin >> n >> a >> b >> c;
    a = 1ll * a * inv(100 - c) % Z;
    b = 1ll * b * inv(100 - c) % Z;

    int fac[2*n];
    fac[0] = 1;
    for (int i = 1; i < 2 * n; i++) {
        fac[i] = 1ll * fac[i-1] * i % Z;
    }

    int ans = 0;
    int ifn1 = inv(fac[n-1]);
    for (int i = 0; i < n; i++) {
        int coe = 1ll * fac[n-1+i] * inv(fac[i]) % Z * ifn1 % Z;
        ans = (ans + 1ll * a * coe % Z * fp(a, n - 1) % Z * fp(b, i) % Z * (n + i)) % Z;
        ans = (ans + 1ll * b * coe % Z * fp(b, n - 1) % Z * fp(a, i) % Z * (n + i)) % Z;
    }

    ans = 1ll * ans * 100 % Z * inv(100 - c) % Z;
    cout << ans << endl;

    return 0;
}
