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>
#define int long long
using namespace std;

int powmod(int a, int n, int mod) {
	if (n == 0) return 1;
	if (n % 2 == 0) return powmod((a * a) % mod, n / 2, mod);
	return (a * powmod(a, n - 1, mod)) % mod;
}

int mod = 1000000007;
int n, a, b, c;
int fact[200001];
int finv[200001];

int comb(int n, int k) {
	return fact[n] * finv[k] % mod * finv[n - k] % mod;
}

signed main() {
	int i;
	
	cin >> n >> a >> b >> c;
	
	fact[0] = 1;
	finv[0] = 1;
	for (i = 1; i < 2 * n; i++) {
		fact[i] = i * fact[i - 1] % mod;
		finv[i] = powmod(fact[i], mod - 2, mod);
	}
	
	int ans = 0;
	int inv = powmod(a + b, mod - 2, mod);
	
	for (i = 0; i < n; i++) {
		int p = powmod(a, n, mod) * powmod(b, i, mod) % mod * powmod(inv, n + i, mod) % mod * comb(n - 1 + i, i) % mod;
		ans += p * (n + i) % mod;
		ans %= mod;
	}
	
	for (i = 0; i < n; i++) {
		int p = powmod(b, n, mod) * powmod(a, i, mod) % mod * powmod(inv, n + i, mod) % mod * comb(n - 1 + i, i) % mod;
		ans += p * (n + i) % mod;
		ans %= mod;
	}
	
	ans = ans * 100 % mod * inv % mod;
	cout << ans << endl;
	return 0;
}