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;
 
#define ll long long
#define all(aaa) aaa.begin(), aaa.end()

const int MOD = 1e9 + 7, N = 2e5 + 5;
ll fac[N], inv_fac[N];

ll bp(ll a, ll p) {
    ll r = 1;
    while (p > 0) {
        if (p & 1)
            r = r * a % MOD;
        p >>= 1;
        a = a * a % MOD;
    }
    return r;
}

ll cb(int n, int k) {
    return fac[n] * inv_fac[n - k] % MOD * inv_fac[k] % MOD;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    inv_fac[0] = fac[0] = 1;

    for (int i = 1; i < N; i++) {
        fac[i] = fac[i - 1] * i % MOD;
        inv_fac[i] = bp(fac[i], MOD - 2);
    }

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

    ll anime = 100 * bp(100 - c, MOD - 2) % MOD,
        pa = a * bp(a + b, MOD - 2) % MOD,
        pb = b * bp(a + b, MOD - 2) % MOD;

    ll ans = 0;

    for (int i = n; i <= 2 * n - 1; i++) {
        for (int j = 0; j < 2; j++) {
            ans = (ans + 
                cb(i - 1, n - 1) 
                * bp(pa, n) % MOD
                * bp(pb, i - n) % MOD 
                * i % MOD 
                * anime) % MOD;

            swap(pa, pb);
        }
    }

    cout << ans << "\n";

    return 0;
}