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>
#include <cassert>

using namespace std;
typedef long long ll;

const ll MOD = 1e9 + 7;
vector<ll> fact;

void init_fact(int n) {
    fact = vector<ll>(n);
    fact[0] = 1;
    for (int i = 1; i < n; i++) fact[i] = fact[i - 1] * i % MOD;
}

ll calc_pow(ll x, ll pow) {
    if (pow == 0) return 1;
    ll a = calc_pow(x, pow / 2);
    a = a * a % MOD;
    if (pow % 2 == 1) a *= x;
    return a % MOD;
}

ll calc_div(ll a, ll b) {
    return a * calc_pow(b, MOD - 2) % MOD;
}

ll calc_comb(ll n, ll r) {
    if (n < r)return 0;
    return calc_div(fact[n], fact[r] * fact[n - r] % MOD);
}

int main() {
    init_fact(250000);

    int n, a, b, c;
    cin >> n >> a >> b >> c;

    ll diva = calc_div(a, a + b), divb = calc_div(b, a + b), divc = calc_div(100, a + b);

    ll ret = 0;
    for (int i = 0; i < n; i++) {
        ll x = calc_pow(diva, n);
        x *= calc_pow(divb, i), x %= MOD;
        ll y = calc_pow(divb, n);
        y *= calc_pow(diva, i), y %= MOD;

        ll now = x + y;
        now *= calc_comb(n + i - 1, n - 1), now %= MOD;
        now *= divc, now %= MOD;
        ret += now * (n + i) % MOD, ret %= MOD;
    }

    cout << ret << endl;
    return 0;
}