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>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

const ll MOD = 1e9 + 7;
const ll max_n = 200010;

ll fact[max_n];
bool fact_init = false;

ll mod(ll a, ll b){
  a %= b;
  return a < 0 ? a + b : a;
}

ll powmod(ll a, ll b){
  if(b == 0) return 1;
  if(b & 1) {
    return a * powmod(a, b - 1) % MOD;
  } else {
    ll d = powmod(a, b / 2) % MOD;
    return d * d % MOD;
  }
}

ll sub(ll a, ll b){
  a %= MOD, b %= MOD;
  a -= b;
  return a < 0 ? a + MOD : a;
}

ll inv(ll a){
	return powmod(a, MOD - 2);
}

ll comb(ll n, ll k){
	if(!fact_init){
		fact[0] = 1;
		fact[1] = 1;

		for(ll i = 2; i < max_n; i++){
			fact[i] = fact[i-1] * i;
			fact[i] %= MOD;
		}

		fact_init = true;
	}

	ll ret = fact[n];
	ret *= powmod(fact[k],MOD-2);
	ret %= MOD;
	ret *= powmod(fact[n-k],MOD-2);
	ret %= MOD;
	return ret;

	// X^(-1) = X^(p-2) (mod p) (Fermat's little theorem)
}
int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

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

	ll awin = a * inv(a + b) % MOD;
    ll bwin = b * inv(a + b) % MOD;
    ll num = 100 * inv(100 - c) % MOD;

    ll ret = 0;
    for(int i = 0; i <= n - 1; i++){
        ret += comb(n - 1 + i, i) * powmod(awin, n) % MOD * powmod(bwin, i) % MOD * (n + i) % MOD;
        ret += comb(n - 1 + i, i) * powmod(bwin, n) % MOD * powmod(awin, i) % MOD * (n + i) % MOD;
    }

    cout << ret % MOD * num % MOD << endl;
}
