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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<int> Vi;
typedef vector<ll> Vl;
typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;


//===

const ll mod = static_cast<ll>(1e9 + 7);




class ModComb {
private:
	Vl fac, inv, finv;
public:
	ModComb(ll n) : fac(n), inv(n), finv(n) {
		fac[0] = fac[1] = 1;
		finv[0] = finv[1] = 1;
		inv[1] = 1;
		for (ll i = 2; i < n; ++i) {
			fac[i] = fac[i - 1] * i % mod;
			inv[i] = mod - inv[mod % i] * (mod / i) % mod;
			finv[i] = finv[i - 1] * inv[i] % mod;
		}
	}

	ll comb(ll n, ll k) {
		if (n < k) return 0;
		if (n < 0 || k < 0) return 0;
		return fac[n] * (finv[n - k] * finv[k] % mod) % mod;
	}
};

ll mpow(ll a, ll n) {
	ll res = 1;
	while (n != 0) {
		if (n & 1) res = res * a % mod;
		n >>= 1;
		a = a * a % mod;
	}
	return res;
}



int main() {
	ll N, A, B, C;
	cin >> N >> A >> B >> C;

	ModComb mc(2 * N);

	A = A * mpow(100 - C, mod - 2) % mod;
	A = A * 100 % mod;
	B = B * mpow(100 - C, mod - 2) % mod;
	B = B * 100 % mod;
	
	ll ans = 0;
	for (ll M = N; M <= 2 * N - 1; ++M) {
		ll tmp, tmp2;
		tmp = mpow(A, N) * mpow(B, M - N) % mod;
		tmp = (tmp + mpow(A, M - N) * mpow(B, N) % mod) % mod;
		tmp = tmp * M % mod;
		tmp = tmp * mc.comb(M - 1, N - 1) % mod;

		tmp2 = mpow(100, M - 1) * (100 - C) % mod;

		ans = ans + (tmp * mpow(tmp2, mod - 2) % mod);
		ans = ans % mod;
	}

	cout << ans << endl;
}