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;

const int mod = 1e9+7;

long long int fac[200010],inv[200010];

long long int mp(long long int a,long long int b){
	a%=mod;
	long long int res=1;
	while(b){
		if(b&1)	res=res*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return res;
}

long long int C(int a,int b){
	return fac[a]*inv[b]%mod*inv[a-b]%mod;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	long long int n,A,B,tmp;
	cin>>n>>A>>B>>tmp;
	fac[0]=1;
	for(int i=1;i<=200000;i++)
		fac[i]=fac[i-1]*i%mod;
	for(int i=0;i<=200000;i++)
		inv[i]=mp(fac[i],mod-2);
	A=A*mp(100-tmp,mod-2)%mod;
	B=B*mp(100-tmp,mod-2)%mod;
	long long int res=0;
	for(int i=0;i<n;i++){
		res=(res+C(n+i-1,n-1)*mp(A,n)%mod*mp(B,i)%mod*(n+i))%mod;
		res=(res+C(n+i-1,n-1)*mp(B,n)%mod*mp(A,i)%mod*(n+i))%mod;
	}
	res=res*100%mod*mp(100-tmp,mod-2)%mod;
	cout<<res<<endl;

}

