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 <iostream>
using namespace std;
#define MOD 1000000007

long powMod(long x, long n, long mod){
	if (n == 0) return 1;
	else if (n == 1) return (x % mod);
	else if (n % 2 == 0){
		long t = powMod(x, n/2, mod);
		return (t*t) % mod;
	}
	else{
		return (x*powMod(x, n-1, mod)) % mod;
	}
}

long divMod(long a, long b, long mod){
	long t = powMod(b, mod-2, mod);
	return (a*t) % mod;
}

long n, a, b, c, p, q, t;
long pn[100010], qn[100010], cn[100010];

int main(){
	cin >> n >> a >> b >> c;
	long out = 0;
	p = divMod(a, a+b, MOD);
	q = divMod(b, a+b, MOD);
	pn[0] = 1;  // p^i
	qn[0] = 1;  // q^i
	cn[0] = 1;  // comb(n+i,n)
	for (int i = 1; i <= n+1; ++i){
		pn[i] = (pn[i-1]*p) % MOD;
		qn[i] = (qn[i-1]*q) % MOD;
		cn[i] = (cn[i-1]*divMod(n+i, i, MOD)) % MOD;
	}
	for (long m = n; m < 2*n; ++m){
		t = (pn[n] * qn[m-n]) % MOD;
		t = (t + (pn[m-n] * qn[n]) % MOD) % MOD;
		t = (cn[m-n]*t) % MOD;
		out = (out + t) % MOD;
	}
	out = (out * divMod(100, 100-c, MOD)) % MOD;
	out = (out * n) % MOD;
	printf("%ld", out);
}