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 <vector>
#include <chrono>
#include <random>

std::mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count());

const int MOD = 1e9 + 7;
const int ms = 200200;

long long fexp(long long x, long long e = MOD - 2) {
	long long ans = 1;
	for(; e > 0; e /= 2) {
		if(e & 1) ans = ans * x % MOD;
		x = x * x % MOD;
	}
	return ans;
}

long long fat[ms], ifat[ms];

long long comb(int n, int a) {
	if(a < 0 || a > n) return 0;
	else return fat[n] * ifat[a] % MOD * ifat[n-a] % MOD;
}

int main() {
	std::ios_base::sync_with_stdio(false); std::cin.tie(NULL);
	int n;
	int a, b, c;
	std::cin >> n >> a >> b >> c;
	int r = fexp((100 - c) * fexp(100) % MOD);
	a = a * fexp(100 - c) % MOD;
	b = b * fexp(100 - c) % MOD;
	long long ans = 0;
	fat[0] = ifat[0] = 1;
	for(int i = 1; i < ms; i++) {
		fat[i] = fat[i-1] * i % MOD;
		ifat[i] = fexp(fat[i]);
	}
	for(int i = 0; i < n; i++) {
		//std::cout << "(" << comb(n + i - 1, i) << " * " << fexp(a, i) << " * " << fexp(b, n) << " * " << r << ")\n";
		ans = (ans + comb(n + i - 1, i) * fexp(a, i) % MOD * fexp(b, n) % MOD * r % MOD * (i + n) % MOD) % MOD;
		ans = (ans + comb(n + i - 1, i) * fexp(b, i) % MOD * fexp(a, n) % MOD * r % MOD * (i + n) % MOD) % MOD;
	}
	std::cout << ans << '\n';
}