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;
const int mod = 1000000007;
long long modpow(long long n, int k) {
	long long r = 1;
	while (k > 0) {
		if (k&1) (r *= n) %= mod;
		(n *= n) %= mod;
		k >>= 1;
	}
	return r;
}
long long modinv(int n) {
	return modpow(n, mod-2);
}
vector<long long> fact;
int main() {
	int N, A, B, C;
	cin >> N >> A >> B >> C;

	fact = vector<long long>(2*N+10);
	fact[0] = 1;
	for (unsigned i = 1; i < fact.size(); ++ i) fact[i] = fact[i-1] * i % mod;

	long long p1 = A * modinv(A+B) % mod;
	long long p2 = B * modinv(A+B) % mod;
	long long sum1 = 0;
	long long sum2 = 0;
	for (int i = 0; i < N; ++ i) {
		long long c = fact[N-1+i] * modinv(fact[i]) % mod * modinv(fact[N-1]) % mod;
		long long x1 = modpow(p2, i) * c % mod;
		long long x2 = modpow(p1, i) * c % mod;
		(x1 *= N+i) %= mod;
		(x2 *= N+i) %= mod;
		(sum1 += x1) %= mod;
		(sum2 += x2) %= mod;
	}
	(sum1 *= modpow(p1, N)) %= mod;
	(sum2 *= modpow(p2, N)) %= mod;
	long long res = (sum1+sum2) % mod;
	res = res * 100 % mod * modinv(A+B) % mod;
	cout << res << endl;
}
