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>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

long long int MOD = 1000000007;

long long int POW(long long int K, long long int p) {

	if (p == 0) { return 1; }
	else if (p == 1) { return K; }

	if ((p % 2) == 1) { return (POW(K, p - 1)*K) % MOD; }

	long long int tmp = POW(K, p / 2);

	return (tmp*tmp) % MOD;
}

long long int INV(long long int K){

	return POW(K, MOD - 2);
}

long long int b[200005];

long long int N, A, B, C;

int main(void) {

	cin >> N >> A >> B >> C;

	b[0] = 1;
	b[1] = 1;

	for (long long int i = 2; i <= 200000; i++) {
		b[i] = b[i - 1] * i;
		b[i] %= MOD;
	}

	long long int ans = 0;

	long long int p, q;
	p = (A*INV(100 - C)) % MOD;
	q = (B*INV(100 - C)) % MOD;


	for (int k = 0; k <= N - 1; k++) {
		long long int fac = b[N + k];
		fac *= INV(b[k]);  fac %= MOD;
		fac *= INV(b[N - 1]); fac %= MOD;
		long long int tmp;
		tmp = fac*((POW(q, k)*POW(p, N) + POW(p, k)*POW(q, N)) % MOD) % MOD;
		ans += tmp;
	}

	ans = (ans * 100) % MOD;
	ans *= INV(100 - C); ans %= MOD;

	cout << ans << endl;




	return 0;
}