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 ll;

const ll MOD = 1e9+7;

ll calc(ll a,ll b){//a^(b) mod  
	if(b==0)return 1;
	ll y = calc(a,b/2);y=(y*y)%MOD;
	if(b & 1) y = (y * a) % MOD;
	return y;
}
ll inv(ll a){
	return calc(a,MOD-2);
}

int main(){

	ll N, A, B, C;
	cin >> N >> A >> B >> C;
	ll ANSWER = 0;
	ll DACOR = 100 * inv(100-C) % MOD;
	vector<ll> a(2*N), b(2*N), f(2*N), fr(2*N);

	a[0] = b[0] = f[0] = fr[0] = f[1] = fr[1] = 1;
	a[1] = A * inv(A+B) % MOD;
	b[1] = B * inv(A+B) % MOD;

	for(int i=2;i<2*N;i++){
		a[i] = a[i-1] * a[1] % MOD;
		b[i] = b[i-1] * b[1] % MOD;
		f[i] = f[i-1] * i % MOD;
		fr[i] = inv(f[i]);
	}

	for(int i=N;i<2*N;i++){
		ll pa = a[N] * b[i-N] % MOD * f[i-1] % MOD * fr[N-1] % MOD * fr[i-N] % MOD;
		ll pb = b[N] * a[i-N] % MOD * f[i-1] % MOD * fr[N-1] % MOD * fr[i-N] % MOD;
		( ANSWER += ( pa + pb ) * i % MOD * DACOR % MOD ) %= MOD;
	}

	cout << ANSWER << endl;
}
