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>

typedef long long ll;
typedef unsigned long long ull;

using namespace std;

const int P=1E9+7;
const int MAXN=2E5+10;

template<typename T1,typename T2>
	void Inc(T1 &a,T2 b){a+=b;if(a>=P) a-=P;}

template<typename T1,typename T2>
	void Dec(T1 &a,T2 b){a-=b;if(a<0) a+=P;}

template<typename T1,typename T2>
	T1 Add(T1 a,T2 b){a+=b;return a>=P?a-P:a;}

template<typename T1,typename T2>
	T1 Sub(T1 a,T2 b){a-=b;return a<0?a+P:a;}

ll ksm(ll a,ll b)
{
	ll ret=1;
	for(;b;b>>=1,(a*=a)%=P)
		if(b&1)
			(ret*=a)%=P;
	return ret;
}

ll fac[MAXN],ifac[MAXN];

ll C(int n,int m)
{
	if(n<m||n<0||m<0)
		return 0;
	return fac[n]*ifac[m]%P*ifac[n-m]%P;
}

int main()
{
	int n,a,b;
	scanf("%d%d%d%*d",&n,&a,&b);
	int c=a+b;
	a=a*ksm(c,P-2)%P;
	b=b*ksm(c,P-2)%P;
	fac[0]=ifac[0]=1;
	for(int i=1;i<MAXN;i++)
		fac[i]=fac[i-1]*i%P;
	ifac[MAXN-1]=ksm(fac[MAXN-1],P-2);
	for(int i=MAXN-2;i;i--)
		ifac[i]=ifac[i+1]*(i+1)%P;
	ll Ans=0;
	for(int i=0;i<n;i++)
		Inc(Ans,ksm(a,n)*ksm(b,i)%P*(i+n)%P*C(n+i-1,i)%P);
	for(int i=0;i<n;i++)
		Inc(Ans,ksm(b,n)*ksm(a,i)%P*(i+n)%P*C(n+i-1,i)%P);
	printf("%lld\n",Ans*100%P*ksm(c,P-2)%P);
}
