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>
#define MOD 1000000007

using namespace std;
typedef long long ll;

vector<ll> d(200000, 0), ten(200000, 0);

ll Expow(ll a, int d) {
	ll ans = 1;
	while (d > 0) {
		if (d & 1)
			ans = ans * a % MOD;
		d >>= 1;
		a = a * a % MOD;
	}
	return ans;
}

ll Inverse(ll a) {
	if (a == 0)
		return 1;
	return Expow(a, MOD - 2);
}

ll Combi(int n, int k) {
	ll a = d[n], b = Inverse(d[k]), c = Inverse(d[n - k]);
	a *= b;
	a %= MOD;
	a *= c;
	a %= MOD;
	return a;
}

void Ini(int n) {
	ll a = 1, t = 100, tt = 100;
	for (int i = 1; i < n; i++) {
		a = a * i % MOD;
		ten[i] = tt;
		tt = tt * t % MOD;
		d[i] = a;
	}
	return;
}
//375000004
int main(void) {
	int n, a, b, c;
	cin >> n >> a >> b >> c;
	if (c != 0) {
		a = a * 100 * Inverse(100 - c) % MOD;
		b = b * 100 * Inverse(100 - c) % MOD;
	}
	Ini(2 * n);
	ll ans, aa, bb, cc, A, B, an = Expow(a, n), bn = Expow(b, n);
	ans = (an + bn) % MOD;
	ans *= Inverse(ten[n]);
	ans %= MOD;
	ans *= n;
	ans %= MOD;
	A = a; B = b;
	for (int i = n; i < 2 * n - 1; i++) {
		cc = Combi(i, n - 1);
		aa = an * B % MOD;
		bb = bn * A % MOD;
		aa = (aa + bb) % MOD;
		aa *= cc;
		aa %= MOD;
		aa *= i + 1;
		aa %= MOD;
		ans += aa * Inverse(ten[i + 1]) % MOD;
		ans %= MOD;
		A = A * a % MOD;
		B = B * b % MOD;
	}
	if (c != 0) {
		ans *= 100;
		ans %= MOD;
		ans *= Inverse(100 - c);
		ans %= MOD;
	}
	cout << ans << "\n";
	return 0;
}