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;

#define ll long long int
#define MOD 1000000007
#define N 200005

ll fact[N], inv[N], finv[N];

ll C(int a, int b) {
    if(a < b || b < 0) return 0;
    return fact[a] * (finv[b] * finv[a - b] % MOD) % MOD;
}

int main() {
    int n;
    ll a, b, c;
    cin >> n >> a >> b >> c;

    fact[0] = 1;
    fact[1] = 1;
    inv[1] = 1;
    finv[1] = finv[0] = 1;
    for(int i = 2; i < N; ++i) {
        fact[i] = fact[i - 1] * i % MOD;
        inv[i] = MOD - (MOD / i) * inv[MOD % i] % MOD;
        finv[i] = finv[i -1] * inv[i] % MOD;
    }

    ll ans = 0, pa = a * inv[a + b] % MOD, pb = b * inv[a + b] % MOD;
    ll powa[N], powb[N];
    powa[0] = 1, powb[0] = 1;

    for(int i = 1;i < N; ++i) {
        powa[i] = powa[i - 1] * pa % MOD;
        powb[i] = powb[i - 1] * pb % MOD;
    }

    for(int i = 0; i < n; ++i) {
        ans = (ans + C(n + i - 1, i) * (i + n) % MOD * ((powa[i] * powb[n] % MOD + powb[i] * powa[n] % MOD))%MOD) % MOD;
    }
    ans = ans * inv[100 - c] % MOD * 100 % MOD;
    cout << ans << endl;

    return 0;
}
