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 lint = long long;
using namespace std;
const int MAXN = 200005;
const int mod = 1e9 + 7;

lint ipow(lint x, lint p){
	lint ret = 1, piv = x;
	while(p){
		if(p & 1) ret = ret * piv % mod;
		piv = piv * piv % mod;
		p >>= 1;
	}
	return ret;
}

lint fact[MAXN], invf[MAXN];

lint bino(int x, int y){
	return fact[x] * (invf[y] * invf[x - y] % mod) % mod;
}

int main(){
	fact[0] = invf[0] = 1;
	for(int i=1; i<MAXN; i++){
		fact[i] = fact[i-1] * i % mod;
		invf[i] = ipow(fact[i], mod - 2);
	}
	int n, a, b, c;
	cin >> n >> a >> b >> c;
	int p = a * ipow(a + b, mod - 2) % mod;
	int q = b * ipow(a + b, mod - 2) % mod; 
	lint ans = 0;
	for(int i=0; i<n; i++){
		int pan = n + i;
		ans += (pan * bino(n - 1 + i, i) % mod) * (ipow(p, n) * ipow(q, i) % mod) % mod;
		ans += (pan * bino(n - 1 + i, i) % mod) * (ipow(q, n) * ipow(p, i) % mod) % mod;
		ans %= mod;
	}
	ans *= ipow(a + b, mod - 2) % mod;
	ans %= mod;
	ans *= 100;
	ans %= mod;
	cout << ans % mod << endl;
}
