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;
const int N = 200000 + 10;
const LL MOD = 1e9 + 7;
int n, a, b, c; 
LL mpow(LL a, LL x) {
	if (x == 0) return 1;
	LL t = mpow(a, x >> 1);
	if (x % 2 == 0) return t * t % MOD;
	return t * t % MOD * a % MOD;
}
LL fac[N], inv[N];
LL C(int x, int y) {
	return fac[x] * inv[y] % MOD * inv[x-y] % MOD;
}
int main() {
	fac[0] = 1; inv[0] = 1;
	for (int i = 1; i < N; i ++) {
		fac[i] = fac[i-1] * i % MOD;
		inv[i] = mpow(fac[i], MOD - 2);
	}
	cin >> n >> a >> b >> c;
	LL p1 = a * mpow(a+b, MOD-2) % MOD;
	LL p2 = b * mpow(a+b, MOD-2) % MOD;
	LL ans = 0;
	for (int i = n; i <= 2 * n - 1; i ++) {
		// ..... A
		LL tmp = C(i-1, n-1) * mpow(p1, n) % MOD * mpow(p2, (i-1)-(n-1)) % MOD; 
		// ..... B
		tmp += C(i-1, n-1) * mpow(p2, n) % MOD * mpow(p1, (i-1)-(n-1)) % MOD;
		tmp %= MOD;
		ans = (ans + tmp * i % MOD) % MOD;
	}
	cout << (ans * 100 % MOD * mpow(a + b, MOD-2) % MOD) << endl;
}
