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;

const int mod = 1e9 + 7;
const int maxn = 200000 + 7;

long long inv[maxn], step[maxn];
long long pow_a[maxn], pow_b[maxn], pow_c[maxn], pow_ab[maxn];
long long inv_step[maxn];

long long C(int n, int m) {
    return step[n] * inv_step[m] % mod * inv_step[n - m] % mod;
}

int main() {
    step[0] = 1;
    for (int i = 1; i < maxn; i++) step[i] = step[i - 1] * i % mod;
    inv[0] = inv[1] = 1;
    for (int i = 2; i < maxn; i++) inv[i] = mod - (mod / i) * inv[mod % i] % mod;
    inv_step[0] = inv[0];
    for (int i = 1; i < maxn; i++) inv_step[i] = inv_step[i - 1] * inv[i] % mod;


    long long n, a, b, c;
    cin >> n >> a >> b >> c;
    pow_a[0] = pow_b[0] = pow_c[0] = pow_ab[0] = 1;
    for (int i = 1; i < maxn; i++) {
        pow_a[i] = pow_a[i - 1] * a % mod;
        pow_b[i] = pow_b[i - 1] * b % mod;
        pow_c[i] = pow_c[i - 1] * c % mod;
        pow_ab[i] = pow_ab[i - 1] * inv[a + b] % mod;
    }

    long long ans = 0;
    for (int i = n; i <= 2 * n - 1; i++) {
        long long tmp = C(i - 1, n - 1);

        tmp = tmp * (pow_a[n] * pow_b[i - n] % mod + pow_a[i - n] * pow_b[n] % mod) % mod * pow_ab[i] % mod;
        tmp = tmp * i % mod * 100 % mod * inv[100 - c] % mod;
        ans = (ans + tmp) % mod;
    }

    cout << ans << endl;
    return 0;
}

