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;

long inv(int a, int m) {
	long i = 1, j = 0;
	for (int b = m; a > 1; swap(i, j), swap(a, b)) {
		j = (j + b / a * (m - i)) % m;
		b %= a;
	}
	return i;
}
long f[222222] = { 1 };
int main() {
	int n, a, b, c, p = 1e9 + 7;
	cin >> n >> a >> b >> c;
	for (int i = 0; i < n * 2; i++)
		f[i + 1] = f[i] * (i + 1) % p;
	long A = 0, B = 0, x = 1, y = 1;
	for (int i = n - 1; i < n * 2 - 1; i++) {
		(A += ((x*f[i]) % p*(i + 1)) % p*inv(f[n - 1] * f[i - n + 1] % p, p)) %= p;
		(x *= b * inv(100 - c, p) % p) %= p;
		(B += ((y*f[i]) % p*(i + 1)) % p*inv(f[n - 1] * f[i - n + 1] % p, p)) %= p;
		(y *= a * inv(100 - c, p) % p) %= p;
	}
	long aa = A * y%p*inv(100 - c, p) % p;
	long bb = B * x%p*inv(100 - c, p) % p;
	cout << (aa + bb) * 100 % p << endl;
	return 0;
}
