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;

constexpr long long MOD = 1e9 + 7;
constexpr int MAX_N = 200000 + 1;
long long fact[MAX_N];
void init_fact() {
    fact[0] = 1;
    for(auto i = 1; i < MAX_N; ++i)
        fact[i] = fact[i-1] * i % MOD;
}

long long pow_mod(long long x, long long y, long long mod) {
    auto ret=1ll;
    while (y) {
            if (y % 2)
                ret = ret * x % mod;
            x =  x * x % mod;
            y /= 2;
        }
    return ret;
}

long long nCr(int n, int r) {
    return fact[n] * pow_mod(fact[n-r], MOD - 2, MOD) % MOD * pow_mod(fact[r], MOD - 2, MOD) % MOD;
}

void solve()
{
    init_fact();
    int N, A, B, C;
    cin >> N >> A >> B >> C;

    long long ans = 0;
    long long dab = pow_mod(A + B, MOD - 2, MOD);
    for(auto i = N; i < 2 * N; ++i)
    {
        ans += nCr(i - 1, N - 1)
            * (pow_mod(A, N, MOD) * pow_mod(B, i - N, MOD) % MOD + pow_mod(A, i - N, MOD) * pow_mod(B, N, MOD) % MOD) % MOD
            * i % MOD 
            * 100 % MOD
            * pow_mod(dab, i, MOD) % MOD
            * pow_mod(100 - C, MOD - 2, MOD) % MOD;
        ans %= MOD;
    }

    cout << ans << endl;
}

int main()
{
    cin.sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    solve();
}