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 maxn=200100,mod=1000000007;
inline void Add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
inline int ksm(int a,int b)
{
	int res=1;
	for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) res=1ll*res*a%mod;
	return res;
}
int fac[maxn],invfac[maxn];
inline int C(int n,int m){return 1ll*fac[n]*invfac[m]%mod*invfac[n-m]%mod;}
int main()
{
	for(int i=fac[0]=1;i<maxn;i++) fac[i]=1ll*fac[i-1]*i%mod;
	invfac[maxn-1]=ksm(fac[maxn-1],mod-2);
	for(int i=maxn-2;i>=0;i--) invfac[i]=1ll*invfac[i+1]*(i+1)%mod;
	int n,a,b,c,ans=0;
	cin>>n>>a>>b>>c;
	for(int cnt=n;cnt<n+n;cnt++)
	{
		Add(ans,1ll*C(cnt-1,n-1)*(1ll*ksm(a,n)*ksm(b,cnt-n)%mod+1ll*ksm(b,n)*ksm(a,cnt-n)%mod)%mod
		*cnt%mod*100%mod*ksm(1ll*ksm(a+b,cnt)*(100-c)%mod,mod-2)%mod);
	}
	cout<<ans<<endl;
}