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>

using namespace std;

typedef long long ll;

const int size = 200005;
const ll mod = 1e9+7;

ll inv[size], fac[size], finv[size];

void comInit() {
		fac[0] = fac[1] = 1;
		finv[0] = finv[1] = 1;
		inv[1] = 1;

		for (int i = 2; i < size; i++) {
				inv[i] = mod - inv[mod%i] * (mod/i) % mod;
				fac[i] = fac[i-1] * (ll)i % mod;
				finv[i] = finv[i-1] * inv[i] % mod;
		}
}

ll com(int a, int b) {
		if (a < b) return 0;
		return fac[a] * (finv[b] * finv[a-b] % mod) % mod;
}

ll modPow(ll x, ll t) {
		if (t == 0) return 1;
		ll res = modPow(x, t/2);
		res = res * res % mod;
		if (t % 2 == 1) res = res * x % mod;
		return res;
}

int main() {
		comInit();
		int N, A, B, C;
		cin >> N >> A >> B >> C;
		ll pa = (ll)A * inv[A+B] % mod;
		ll pb = (ll)B * inv[B+A] % mod;
		ll ans = 0;

		for (int k = 0; k < N; k++) {
				ans += com(k+N-1, k) * modPow(pb, k) % mod * modPow(pa, N) % mod * (ll)(k+N) % mod;
				ans += com(k+N-1, k) * modPow(pa, k) % mod * modPow(pb, N) % mod * (ll)(k+N) % mod;
				ans %= mod;
		}

		ans = ans * inv[A+B] % mod * 100LL % mod;
		cout << ans << endl;
		return 0;
}
