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 REP(x,y,z) for(int x=y;x<=z;x++)
#define MSET(x,y) memset(x,y,sizeof(x))
#define M 200005
#define MOD 1000000007
using namespace std;
using LL = long long;
LL pw(LL x,LL y) {
    LL res=1, tmp=x;
    while (y) {
        if (y&1) res = res*tmp%MOD;
        tmp = tmp*tmp%MOD;
        y /= 2;
    }
    return res;
}
LL fact[M];
LL comb(LL x,LL y) {
    return fact[x] * pw(fact[y], MOD-2) % MOD * pw(fact[x-y], MOD-2) % MOD;
}
int main()
{
    fact[0] = 1;
    REP(i,1,M-1) fact[i] = fact[i-1] * i % MOD;


    int n,a,b,c;
    while (~scanf("%d %d %d %d", &n,&a,&b,&c)) {
        LL ans = 0;

        REP(i,n,2*n-1) {
            int x = n-1;
            int y = i-1-x;
            LL suc = a * pw(a+b, MOD-2) % MOD;
            LL fail = b * pw(a+b, MOD-2) % MOD;
            LL prob = comb(x+y, x) * pw(suc, x+1) % MOD * pw(fail, y) % MOD;
            ans += prob*i;
            ans %= MOD;

            suc = b * pw(a+b, MOD-2) % MOD;
            fail = a * pw(a+b, MOD-2) % MOD;
            prob = comb(x+y, x) * pw(suc, x+1) % MOD * pw(fail, y) % MOD;
            ans += prob*i;
            ans %= MOD;
        }

        ans = ans * 100 % MOD * pw(100-c, MOD-2) % MOD;
        printf("%lld\n", ans);
    }
    return 0;
}
