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 <cstdio>
#include <iostream>
#define mod 1000000007
#define MN 201000

int n, a, b, c;
int fac[MN], inv[MN];

int qpow(int x, int p)
{
	int ans = 1;
	while(p)
	{
		if(p & 1) ans = 1ll * ans * x % mod;
		x = 1ll * x * x % mod; p >>= 1;
	}
	return ans;
}

int C(int a, int b) {return 1ll * fac[a] * inv[b] % mod * inv[a - b] % mod;}

int main()
{
	scanf("%d%d%d%d", &n, &a, &b, &c);

	int N = 2 * n; fac[0] = 1;
	for(int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
	inv[N] = qpow(fac[N], mod - 2);
	for(int i = N - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;

	int P = qpow(a + b, mod - 2);
	a = 1ll * a * P % mod;
	b = 1ll * b * P % mod;
	int w = 100ll * qpow(100 - c, mod - 2) % mod;
	
	int ans = 0;
	for(int i = n; i < N; i++)
	{
		int p1 = 1ll * qpow(a, n) * qpow(b, i - n) % mod;
		ans = (ans + 1ll * p1 * C(i - 1, n - 1) % mod * i) % mod;
		
		int p2 = 1ll * qpow(a, i - n) * qpow(b, n) % mod;
		ans = (ans + 1ll * p2 * C(i - 1, n - 1) % mod * i) % mod;
	}
	
	printf("%lld\n", 1ll * ans * w % mod);
} 