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;
typedef long long int ll;
static const ll MOD = 1000000007;

int N;
ll A, B, C;

ll fact[200001], invfact[200001];

ll mod_pow(ll x, ll n){
	if(n == 0) return 1;
	ll ret = mod_pow(x * x % MOD, n / 2) % MOD;
	if(n & 1) return ret * x % MOD;
	return ret;
}

ll nCk(int n, int k){
	return fact[n] * invfact[n - k] % MOD * invfact[k] % MOD;
}
ll gcd(ll a, ll b){
	if(b == 0) return a;
	return gcd(b, a % b);
}

int main(){
	cin >> N >> A >> B >> C;
	for(int i = 0; i <= 2 * N; i++){
		if(i == 0) fact[i] = 1;
		else{
			fact[i] = fact[i - 1] * i;
			fact[i] %= MOD;
		}
	}
	for(int i = 0; i <= 2 * N; i++){
		if(i == 0) invfact[i] = 1;
		else{
			invfact[i] = invfact[i - 1] * mod_pow(i, MOD - 2);
			invfact[i] %= MOD;
		}
	}
	ll a = A * mod_pow(A + B, MOD - 2) % MOD;
	ll b = B * mod_pow(A + B, MOD - 2) % MOD;
	ll res = 0;
	for(int i = 0; i < N; i++){
		ll p = mod_pow(a, N) * mod_pow(b, i) % MOD;
		p = p * nCk(N - 1 + i, i) % MOD;
		p = p * (N + i) % MOD;
		p = p * 100 % MOD;
		p = p * mod_pow(100 - C, MOD - 2) % MOD;
		res += p;
		res %= MOD;
		ll q = mod_pow(a, i) * mod_pow(b, N) % MOD;
		q = q * nCk(N - 1 + i, i) % MOD;
		q = q * (N + i) % MOD;
		q = q * 100 % MOD;
		q = q * mod_pow(100 - C, MOD - 2) % MOD;
		res += q;
		res %= MOD;
	}
	cout << res << endl;
	return 0;
}
