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 <stdio.h>

#define MOD 1000000007
long long powmod(long long b,int e)
{
    if (e==0) return 1;
    if (e&1) return powmod(b*b%MOD,e/2)*b%MOD;
    return powmod(b*b%MOD,e/2);
}
long long inv[200005];

long long fac[200005];
long long ifac[200005];
long long choose(int a,int b)
{
    if (b<0 || b>a) return 0;
    return (fac[a]*ifac[b]%MOD)*ifac[a-b]%MOD;
}

int n,a,b,c;
long long ans=0;
long long tmp;

int main()
{
    #ifdef NOT_DMOJ
    freopen("data.txt","r",stdin);
    #endif // NOT_DMOJ
    for (int i=1; i<=200000; i++) inv[i]=powmod(i,MOD-2);
    fac[0]=ifac[0]=1;
    for (int i=1; i<=200000; i++) {
        fac[i]=fac[i-1]*i%MOD;
        ifac[i]=ifac[i-1]*inv[i]%MOD;
    }
    scanf("%d%d%d%d",&n,&a,&b,&c);
    for (int i=0; i<n; i++) {
        // play to (n-1,i) then A wins
        tmp=powmod(inv[a+b],n+i); tmp%=MOD;
        tmp*=powmod(a,n); tmp%=MOD;
        tmp*=powmod(b,i); tmp%=MOD;
        tmp*=choose(n+i-1,i); tmp%=MOD;
        //printf("got %d^%d %d^%d = %lld\n",a,n,b,i,tmp);
        ans+=tmp*(n+i); ans%=MOD;
        // play to (i,n-1) then B wins
        tmp=powmod(inv[a+b],n+i); tmp%=MOD;
        tmp*=powmod(b,n); tmp%=MOD;
        tmp*=powmod(a,i); tmp%=MOD;
        tmp*=choose(n+i-1,i); tmp%=MOD;
        ans+=tmp*(n+i); ans%=MOD;
    }
    // consider c
    // ans becomes ans*100/(100-c)
    ans*=100; ans%=MOD;
    ans*=inv[100-c]; ans%=MOD;
    printf("%lld\n",ans);
}
