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>
#define fi first
#define se second
#define bug cout<<"--------------"<<endl
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=1e9;
const ll llf=1e18;
const int mod=1e9+7;
const int maxn=2e5+10;
ll n;
ll a,b,c;
ll f[maxn];
ll ksm(ll a,ll b){
	ll ret=1;
	while(b){
		if(b&1) ret=ret*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ret;
}
ll ks(ll a,ll b,ll mod){
	ll ret=1;
	while(b){
		if(b&1) ret=ret*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ret;
}
ll C(ll n,ll m){
	return f[n]*ksm(f[m],mod-2)%mod*ksm(f[n-m],mod-2)%mod;
}
ll phi(ll n)
{
     ll i,rea=n;
     for(i=2;i*i<=n;i++)
     {
         if(n%i==0)
         {
             rea=rea-rea/i;
             while(n%i==0)
                 n/=i;
          }
     }
     if(n>1)
         rea=rea-rea/n;
     return rea;
}
int main(){
	//ios::sync_with_stdio(false);
	//freopen("in","r",stdin);
	f[0]=1;
	for(int i=1;i<=2e5+5;i++){
		f[i]=f[i-1]*i%mod;
	}
	cin>>n>>a>>b>>c;
	a=a*ksm((a+b),mod-2)%mod;
	b=(1-a+mod)%mod;
	c=c*ksm(100,mod-2)%mod;
	c=(1-c+mod)%mod;
	ll ans=0;
	for(ll i=n;i<=2*n-1;i++){
		ll k=i; 
		ans+=(k*ksm(a,n)%mod*ksm(b,k-n)%mod*C(k-1,n-1)%mod+k*ksm(b,n)%mod*ksm(a,k-n)%mod*C(k-1,n-1)%mod)%mod;
		ans%=mod;
	}
	cout<<ans*ksm(c,mod-2)%mod;
	return 0;
}