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 int long long
using namespace std;
const int N=1e5,P=1e9+7;
int f[N+N+10],inv[N+N+10];
inline int qpow(int a,int b=P-2){
    int res=1;
    for(;b;b>>=1){
        if(b&1)res=1ll*res*a%P;
        a=1ll*a*a%P;
    }
    return res;
}
inline int C(int n,int m){
    //cout<<"????"<<n<<':'<<m<<' '<<f[n]<<' '<<inv[m]<<' '<<inv[n-m]<<endl;
    return 1ll*f[n]*inv[m]%P*inv[n-m]%P;
}
signed main(){
    f[0]=1;
    for(int i=1;i<=N+N;i++){
        f[i]=1ll*f[i-1]*i%P;
    }
    inv[2*N]=qpow(f[2*N]);
    inv[0]=1;
    for(int i=N+N-1;i>=1;i--){
        inv[i]=1ll*(i+1)*inv[i+1]%P;
    }
    int n,a,b,c;
    cin>>n>>a>>b>>c;
    int ti=qpow(100,P-2);
    a=1ll*a*ti%P;
    b=1ll*b*ti%P;
    c=1ll*c*ti%P;
    int ans=0;
    int p1=qpow(a,n),p2=qpow(b,n);
    int c1=1,c2=1;
    c=(P+1-c);    
    if(c>=P)c-=P;
    c=qpow(c);
    int c3=qpow(c,n+1);
    for(int i=n;i<n+n;i++){
        ans=(ans+(1ll*p1*c1%P+1ll*p2*c2%P)*C(i-1,n-1)%P*c3%P*i%P)%P;
        c1=1ll*c1*b%P;
        c2=1ll*c2*a%P;
        c3=1ll*c3*c%P;
    }
    cout<<ans;
}