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<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const ll MOD=1e9+7;
ll a,b,c,fact[1000005],inv[1000005];
int n;
ll quick_pow(ll x,ll a)
{
	ll ans=1;
	while(a)
	{
		if(a&1)ans=ans*x%MOD;
		x=x*x%MOD;
		a>>=1;
	}
	return ans;
}
ll C(int n,int m)
{
	return fact[n]*inv[m]%MOD*inv[n-m]%MOD;
}
int main()
{
	scanf("%d%lld%lld%lld",&n,&a,&b,&c);
	fact[0]=1;
	for(int i=1;i<=1000000;i++)fact[i]=fact[i-1]*i%MOD;
	inv[1000000]=quick_pow(fact[1000000],MOD-2);
	for(int i=1000000;i>=1;i--)inv[i-1]=inv[i]*i%MOD;
	ll pos=(a+b)*quick_pow(100,MOD-2)%MOD;
	ll inv=quick_pow(a+b,MOD-2);
	a=a*inv%MOD,b=b*inv%MOD;
	ll ans=0;
	for(int i=n;i<=2*n-1;i++)
	{
		ans+=C(i-1,n-1)*quick_pow(a,n-1)%MOD*quick_pow(b,i-n)%MOD*a%MOD*i%MOD;
		ans+=C(i-1,n-1)*quick_pow(b,n-1)%MOD*quick_pow(a,i-n)%MOD*b%MOD*i%MOD;
	}
	ans%=MOD;
	printf("%lld\n",ans*quick_pow(pos,MOD-2)%MOD);
	return 0;
}