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
#!/usr/bin/env python3
import sys
from collections import deque, Counter
from heapq import heappop, heappush
from bisect import bisect_right
from itertools import accumulate

p = 10**9 + 7

N, A, B, C = map(int, input().split())
inv100 = pow(100,p-2,p)
A = A * pow(100-C,p-2,p) % p
B = B * pow(100-C,p-2,p) % p
C = C * inv100 % p
# print(A,B,(1-C)%p)

a = [None] * (2*N+1)
inva = [None] * (2*N+1)
a[0] = 1

for i in range(1, 2*N+1):
    a[i] = i * a[i-1] % p

inva[2*N] = pow(a[2*N],p-2,p)
for i in range(2*N):
    inva[2*N-i-1] = inva[2*N-i] * (2*N-i) % p

ans = 0
prevN = pow(A,N,p) * pow(pow(B,N,p),p-2,p) % p# (A/B)^N

for m in range(N,2*N):
    tmp = (a[m-1] * inva[N-1] % p) * inva[m-N] % p
    # tmp = tmp * ((pow(B,m,p)*prevN % p + pow(A,m,p)*pow(prevN,p-2,p) % p) % p) % p
    tmp *= m
    tmp %= p
    tmp *= ((pow(A,m-N,p) * pow(B,N,p) % p) + (pow(B,m-N,p) * pow(A,N,p) % p)) % p
    tmp %= p
    ans += tmp
    ans %= p
print(ans * pow(1-C,p-2,p)%p)